Inequality $frac{a^2}{b}+frac{b^2}{c}+frac{c^2}{a}+15sqrt[3]{abc}geq 6(a+b+c).$












2














Let $a,b$ and $c$ are positive numbers. Prove that
$$dfrac{a^2}{b}+dfrac{b^2}{c}+dfrac{c^2}{a}+15sqrt[3]{abc}geq 6(a+b+c).$$
I tried BW, SOS-Schur, but without any success.



I let $abc=1$. Then assuming $a=y/x,b=z/y,c=x/z$.



I got a six degree polynomial. But , I failed to complete the proof.






inequality buffalo-way







share|cite|improve this question
























  • I know that you are strong enough in inequalities.What is your trying? Show please. Otherwise, I can't show to you my solution.
    – Michael Rozenberg
    Nov 6 '18 at 5:37












  • I tried BW, SOS-Schur, but without any success.
    – yao4015
    Nov 6 '18 at 5:42










  • BW helps here, but I think there is something more nicer.
    – Michael Rozenberg
    Nov 6 '18 at 5:58
















2














Let $a,b$ and $c$ are positive numbers. Prove that
$$dfrac{a^2}{b}+dfrac{b^2}{c}+dfrac{c^2}{a}+15sqrt[3]{abc}geq 6(a+b+c).$$
I tried BW, SOS-Schur, but without any success.



I let $abc=1$. Then assuming $a=y/x,b=z/y,c=x/z$.



I got a six degree polynomial. But , I failed to complete the proof.






inequality buffalo-way







share|cite|improve this question
























  • I know that you are strong enough in inequalities.What is your trying? Show please. Otherwise, I can't show to you my solution.
    – Michael Rozenberg
    Nov 6 '18 at 5:37












  • I tried BW, SOS-Schur, but without any success.
    – yao4015
    Nov 6 '18 at 5:42










  • BW helps here, but I think there is something more nicer.
    – Michael Rozenberg
    Nov 6 '18 at 5:58














2












2








2


2





Let $a,b$ and $c$ are positive numbers. Prove that
$$dfrac{a^2}{b}+dfrac{b^2}{c}+dfrac{c^2}{a}+15sqrt[3]{abc}geq 6(a+b+c).$$
I tried BW, SOS-Schur, but without any success.



I let $abc=1$. Then assuming $a=y/x,b=z/y,c=x/z$.



I got a six degree polynomial. But , I failed to complete the proof.






inequality buffalo-way







share|cite|improve this question















Let $a,b$ and $c$ are positive numbers. Prove that
$$dfrac{a^2}{b}+dfrac{b^2}{c}+dfrac{c^2}{a}+15sqrt[3]{abc}geq 6(a+b+c).$$
I tried BW, SOS-Schur, but without any success.



I let $abc=1$. Then assuming $a=y/x,b=z/y,c=x/z$.



I got a six degree polynomial. But , I failed to complete the proof.






inequality buffalo-way




inequality buffalo-way






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 12:20









Martin Sleziak

44.7k7115270




44.7k7115270










asked Nov 6 '18 at 5:34









yao4015

655




655












  • I know that you are strong enough in inequalities.What is your trying? Show please. Otherwise, I can't show to you my solution.
    – Michael Rozenberg
    Nov 6 '18 at 5:37












  • I tried BW, SOS-Schur, but without any success.
    – yao4015
    Nov 6 '18 at 5:42










  • BW helps here, but I think there is something more nicer.
    – Michael Rozenberg
    Nov 6 '18 at 5:58


















  • I know that you are strong enough in inequalities.What is your trying? Show please. Otherwise, I can't show to you my solution.
    – Michael Rozenberg
    Nov 6 '18 at 5:37












  • I tried BW, SOS-Schur, but without any success.
    – yao4015
    Nov 6 '18 at 5:42










  • BW helps here, but I think there is something more nicer.
    – Michael Rozenberg
    Nov 6 '18 at 5:58
















I know that you are strong enough in inequalities.What is your trying? Show please. Otherwise, I can't show to you my solution.
– Michael Rozenberg
Nov 6 '18 at 5:37






I know that you are strong enough in inequalities.What is your trying? Show please. Otherwise, I can't show to you my solution.
– Michael Rozenberg
Nov 6 '18 at 5:37














I tried BW, SOS-Schur, but without any success.
– yao4015
Nov 6 '18 at 5:42




I tried BW, SOS-Schur, but without any success.
– yao4015
Nov 6 '18 at 5:42












BW helps here, but I think there is something more nicer.
– Michael Rozenberg
Nov 6 '18 at 5:58




BW helps here, but I think there is something more nicer.
– Michael Rozenberg
Nov 6 '18 at 5:58










1 Answer
1






active

oldest

votes


















2














BW helps here!



We need to prove that
$$frac{a^6}{b^3}+frac{b^6}{c^3}+frac{c^6}{a^3}+15abcgeq6(a^3+b^3+c^3).$$
Now, let $a=min{a,b,c}$, $b=a+u$ and $c=a+v$.



Thus, we need to prove that:
$$3(u^2-uv+v^2)a^{10}+(13u^3-45u^2v+36uv^2+13v^3)a^9+$$
$$+3(17u^4-55u^3v+18u^2v^2+26uv^3+17v^4)a^8+$$
$$+6(15u^5-35u^4v-12u^2v^2+24u^2v^3+28uv^4+15v^5)a^7+$$
$$+6(13u^6-18u^5v-30u^4v^2+13u^3v^3+33u^2v^4+45uv^5+13v^6)a^6+$$
$$+6(6u^7-3u^6v-18u^5v^2-5u^4v^3+16u^3v^4+45u^2v^5+39uv^6+6v^7)a^5+$$
$$+3(3u^8-6u^6v^2-12u^5v^3+5u^4v^4+30u^3v^5+78u^2v^6+36uv^7+3u^8)a^4+$$
$$+(u^9-6u^6v^3+78u^3v^6+108u^2v^7+27uv^8+v^9)a^3+$$
$$+3(12u^2+9uv+v^2)uv^7a^2+3u^2v^8a(3u+v)+u^3v^9geq0,$$ which is true because
$$3(u^2-uv+v^2)geq3uv,$$
$$13u^3-45u^2v+36uv^2+13v^3geq3sqrt{u^3v^3},$$
$$3(17u^4-55u^3v+18u^2v^2+26uv^3+17v^4)geq-21u^2v^2,$$
$$6(15u^5-35u^4v-12u^2v^2+24u^2v^3+28uv^4+15v^5)geq-16sqrt{u^5v^5},$$
$$6(13u^6-18u^5v-30u^4v^2+13u^3v^3+33u^2v^4+45uv^5+13v^6)geq65u^3v^3,$$
$$6(6u^7-3u^6v-18u^5v^2-5u^4v^3+16u^3v^4+45u^2v^5+39uv^6+6v^7)geq144sqrt{u^7v^7},$$
$$3(3u^8-6u^6v^2-12u^5v^3+5u^4v^4+30u^3v^5+78u^2v^6+36uv^7+3u^8)geq127u^4v^4$$ and
$$u^9-6u^6v^3+78u^3v^6+108u^2v^7+27uv^8+v^9geq88sqrt{u^9v^9}.$$
Now, let $frac{a}{sqrt{uv}}=x$.



Thus, it's enough to prove that
$$3x^7+3x^6-21x^5-16x^4+65x^3+144x^2+127x+88geq0,$$ which is obvious.






share|cite|improve this answer























  • Firstly, I let $abc=1$ . Then assuming $ a=y/x, b=z/y,c=x/z$ .I got a six degree polynomial. But , I failed to complete the proof. I want to see more beautiful proofs.
    – yao4015
    Nov 6 '18 at 8:33











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1 Answer
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active

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active

oldest

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active

oldest

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2














BW helps here!



We need to prove that
$$frac{a^6}{b^3}+frac{b^6}{c^3}+frac{c^6}{a^3}+15abcgeq6(a^3+b^3+c^3).$$
Now, let $a=min{a,b,c}$, $b=a+u$ and $c=a+v$.



Thus, we need to prove that:
$$3(u^2-uv+v^2)a^{10}+(13u^3-45u^2v+36uv^2+13v^3)a^9+$$
$$+3(17u^4-55u^3v+18u^2v^2+26uv^3+17v^4)a^8+$$
$$+6(15u^5-35u^4v-12u^2v^2+24u^2v^3+28uv^4+15v^5)a^7+$$
$$+6(13u^6-18u^5v-30u^4v^2+13u^3v^3+33u^2v^4+45uv^5+13v^6)a^6+$$
$$+6(6u^7-3u^6v-18u^5v^2-5u^4v^3+16u^3v^4+45u^2v^5+39uv^6+6v^7)a^5+$$
$$+3(3u^8-6u^6v^2-12u^5v^3+5u^4v^4+30u^3v^5+78u^2v^6+36uv^7+3u^8)a^4+$$
$$+(u^9-6u^6v^3+78u^3v^6+108u^2v^7+27uv^8+v^9)a^3+$$
$$+3(12u^2+9uv+v^2)uv^7a^2+3u^2v^8a(3u+v)+u^3v^9geq0,$$ which is true because
$$3(u^2-uv+v^2)geq3uv,$$
$$13u^3-45u^2v+36uv^2+13v^3geq3sqrt{u^3v^3},$$
$$3(17u^4-55u^3v+18u^2v^2+26uv^3+17v^4)geq-21u^2v^2,$$
$$6(15u^5-35u^4v-12u^2v^2+24u^2v^3+28uv^4+15v^5)geq-16sqrt{u^5v^5},$$
$$6(13u^6-18u^5v-30u^4v^2+13u^3v^3+33u^2v^4+45uv^5+13v^6)geq65u^3v^3,$$
$$6(6u^7-3u^6v-18u^5v^2-5u^4v^3+16u^3v^4+45u^2v^5+39uv^6+6v^7)geq144sqrt{u^7v^7},$$
$$3(3u^8-6u^6v^2-12u^5v^3+5u^4v^4+30u^3v^5+78u^2v^6+36uv^7+3u^8)geq127u^4v^4$$ and
$$u^9-6u^6v^3+78u^3v^6+108u^2v^7+27uv^8+v^9geq88sqrt{u^9v^9}.$$
Now, let $frac{a}{sqrt{uv}}=x$.



Thus, it's enough to prove that
$$3x^7+3x^6-21x^5-16x^4+65x^3+144x^2+127x+88geq0,$$ which is obvious.






share|cite|improve this answer























  • Firstly, I let $abc=1$ . Then assuming $ a=y/x, b=z/y,c=x/z$ .I got a six degree polynomial. But , I failed to complete the proof. I want to see more beautiful proofs.
    – yao4015
    Nov 6 '18 at 8:33
















2














BW helps here!



We need to prove that
$$frac{a^6}{b^3}+frac{b^6}{c^3}+frac{c^6}{a^3}+15abcgeq6(a^3+b^3+c^3).$$
Now, let $a=min{a,b,c}$, $b=a+u$ and $c=a+v$.



Thus, we need to prove that:
$$3(u^2-uv+v^2)a^{10}+(13u^3-45u^2v+36uv^2+13v^3)a^9+$$
$$+3(17u^4-55u^3v+18u^2v^2+26uv^3+17v^4)a^8+$$
$$+6(15u^5-35u^4v-12u^2v^2+24u^2v^3+28uv^4+15v^5)a^7+$$
$$+6(13u^6-18u^5v-30u^4v^2+13u^3v^3+33u^2v^4+45uv^5+13v^6)a^6+$$
$$+6(6u^7-3u^6v-18u^5v^2-5u^4v^3+16u^3v^4+45u^2v^5+39uv^6+6v^7)a^5+$$
$$+3(3u^8-6u^6v^2-12u^5v^3+5u^4v^4+30u^3v^5+78u^2v^6+36uv^7+3u^8)a^4+$$
$$+(u^9-6u^6v^3+78u^3v^6+108u^2v^7+27uv^8+v^9)a^3+$$
$$+3(12u^2+9uv+v^2)uv^7a^2+3u^2v^8a(3u+v)+u^3v^9geq0,$$ which is true because
$$3(u^2-uv+v^2)geq3uv,$$
$$13u^3-45u^2v+36uv^2+13v^3geq3sqrt{u^3v^3},$$
$$3(17u^4-55u^3v+18u^2v^2+26uv^3+17v^4)geq-21u^2v^2,$$
$$6(15u^5-35u^4v-12u^2v^2+24u^2v^3+28uv^4+15v^5)geq-16sqrt{u^5v^5},$$
$$6(13u^6-18u^5v-30u^4v^2+13u^3v^3+33u^2v^4+45uv^5+13v^6)geq65u^3v^3,$$
$$6(6u^7-3u^6v-18u^5v^2-5u^4v^3+16u^3v^4+45u^2v^5+39uv^6+6v^7)geq144sqrt{u^7v^7},$$
$$3(3u^8-6u^6v^2-12u^5v^3+5u^4v^4+30u^3v^5+78u^2v^6+36uv^7+3u^8)geq127u^4v^4$$ and
$$u^9-6u^6v^3+78u^3v^6+108u^2v^7+27uv^8+v^9geq88sqrt{u^9v^9}.$$
Now, let $frac{a}{sqrt{uv}}=x$.



Thus, it's enough to prove that
$$3x^7+3x^6-21x^5-16x^4+65x^3+144x^2+127x+88geq0,$$ which is obvious.






share|cite|improve this answer























  • Firstly, I let $abc=1$ . Then assuming $ a=y/x, b=z/y,c=x/z$ .I got a six degree polynomial. But , I failed to complete the proof. I want to see more beautiful proofs.
    – yao4015
    Nov 6 '18 at 8:33














2












2








2






BW helps here!



We need to prove that
$$frac{a^6}{b^3}+frac{b^6}{c^3}+frac{c^6}{a^3}+15abcgeq6(a^3+b^3+c^3).$$
Now, let $a=min{a,b,c}$, $b=a+u$ and $c=a+v$.



Thus, we need to prove that:
$$3(u^2-uv+v^2)a^{10}+(13u^3-45u^2v+36uv^2+13v^3)a^9+$$
$$+3(17u^4-55u^3v+18u^2v^2+26uv^3+17v^4)a^8+$$
$$+6(15u^5-35u^4v-12u^2v^2+24u^2v^3+28uv^4+15v^5)a^7+$$
$$+6(13u^6-18u^5v-30u^4v^2+13u^3v^3+33u^2v^4+45uv^5+13v^6)a^6+$$
$$+6(6u^7-3u^6v-18u^5v^2-5u^4v^3+16u^3v^4+45u^2v^5+39uv^6+6v^7)a^5+$$
$$+3(3u^8-6u^6v^2-12u^5v^3+5u^4v^4+30u^3v^5+78u^2v^6+36uv^7+3u^8)a^4+$$
$$+(u^9-6u^6v^3+78u^3v^6+108u^2v^7+27uv^8+v^9)a^3+$$
$$+3(12u^2+9uv+v^2)uv^7a^2+3u^2v^8a(3u+v)+u^3v^9geq0,$$ which is true because
$$3(u^2-uv+v^2)geq3uv,$$
$$13u^3-45u^2v+36uv^2+13v^3geq3sqrt{u^3v^3},$$
$$3(17u^4-55u^3v+18u^2v^2+26uv^3+17v^4)geq-21u^2v^2,$$
$$6(15u^5-35u^4v-12u^2v^2+24u^2v^3+28uv^4+15v^5)geq-16sqrt{u^5v^5},$$
$$6(13u^6-18u^5v-30u^4v^2+13u^3v^3+33u^2v^4+45uv^5+13v^6)geq65u^3v^3,$$
$$6(6u^7-3u^6v-18u^5v^2-5u^4v^3+16u^3v^4+45u^2v^5+39uv^6+6v^7)geq144sqrt{u^7v^7},$$
$$3(3u^8-6u^6v^2-12u^5v^3+5u^4v^4+30u^3v^5+78u^2v^6+36uv^7+3u^8)geq127u^4v^4$$ and
$$u^9-6u^6v^3+78u^3v^6+108u^2v^7+27uv^8+v^9geq88sqrt{u^9v^9}.$$
Now, let $frac{a}{sqrt{uv}}=x$.



Thus, it's enough to prove that
$$3x^7+3x^6-21x^5-16x^4+65x^3+144x^2+127x+88geq0,$$ which is obvious.






share|cite|improve this answer














BW helps here!



We need to prove that
$$frac{a^6}{b^3}+frac{b^6}{c^3}+frac{c^6}{a^3}+15abcgeq6(a^3+b^3+c^3).$$
Now, let $a=min{a,b,c}$, $b=a+u$ and $c=a+v$.



Thus, we need to prove that:
$$3(u^2-uv+v^2)a^{10}+(13u^3-45u^2v+36uv^2+13v^3)a^9+$$
$$+3(17u^4-55u^3v+18u^2v^2+26uv^3+17v^4)a^8+$$
$$+6(15u^5-35u^4v-12u^2v^2+24u^2v^3+28uv^4+15v^5)a^7+$$
$$+6(13u^6-18u^5v-30u^4v^2+13u^3v^3+33u^2v^4+45uv^5+13v^6)a^6+$$
$$+6(6u^7-3u^6v-18u^5v^2-5u^4v^3+16u^3v^4+45u^2v^5+39uv^6+6v^7)a^5+$$
$$+3(3u^8-6u^6v^2-12u^5v^3+5u^4v^4+30u^3v^5+78u^2v^6+36uv^7+3u^8)a^4+$$
$$+(u^9-6u^6v^3+78u^3v^6+108u^2v^7+27uv^8+v^9)a^3+$$
$$+3(12u^2+9uv+v^2)uv^7a^2+3u^2v^8a(3u+v)+u^3v^9geq0,$$ which is true because
$$3(u^2-uv+v^2)geq3uv,$$
$$13u^3-45u^2v+36uv^2+13v^3geq3sqrt{u^3v^3},$$
$$3(17u^4-55u^3v+18u^2v^2+26uv^3+17v^4)geq-21u^2v^2,$$
$$6(15u^5-35u^4v-12u^2v^2+24u^2v^3+28uv^4+15v^5)geq-16sqrt{u^5v^5},$$
$$6(13u^6-18u^5v-30u^4v^2+13u^3v^3+33u^2v^4+45uv^5+13v^6)geq65u^3v^3,$$
$$6(6u^7-3u^6v-18u^5v^2-5u^4v^3+16u^3v^4+45u^2v^5+39uv^6+6v^7)geq144sqrt{u^7v^7},$$
$$3(3u^8-6u^6v^2-12u^5v^3+5u^4v^4+30u^3v^5+78u^2v^6+36uv^7+3u^8)geq127u^4v^4$$ and
$$u^9-6u^6v^3+78u^3v^6+108u^2v^7+27uv^8+v^9geq88sqrt{u^9v^9}.$$
Now, let $frac{a}{sqrt{uv}}=x$.



Thus, it's enough to prove that
$$3x^7+3x^6-21x^5-16x^4+65x^3+144x^2+127x+88geq0,$$ which is obvious.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 6 '18 at 6:29

























answered Nov 6 '18 at 6:12









Michael Rozenberg

96.2k1588186




96.2k1588186












  • Firstly, I let $abc=1$ . Then assuming $ a=y/x, b=z/y,c=x/z$ .I got a six degree polynomial. But , I failed to complete the proof. I want to see more beautiful proofs.
    – yao4015
    Nov 6 '18 at 8:33


















  • Firstly, I let $abc=1$ . Then assuming $ a=y/x, b=z/y,c=x/z$ .I got a six degree polynomial. But , I failed to complete the proof. I want to see more beautiful proofs.
    – yao4015
    Nov 6 '18 at 8:33
















Firstly, I let $abc=1$ . Then assuming $ a=y/x, b=z/y,c=x/z$ .I got a six degree polynomial. But , I failed to complete the proof. I want to see more beautiful proofs.
– yao4015
Nov 6 '18 at 8:33




Firstly, I let $abc=1$ . Then assuming $ a=y/x, b=z/y,c=x/z$ .I got a six degree polynomial. But , I failed to complete the proof. I want to see more beautiful proofs.
– yao4015
Nov 6 '18 at 8:33


















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