Proof verification: For $a$, $b$, $c$ positive with $abc=1$, show $sum_{text{cyc}}frac{1}{a^3(b+c)}geq...
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I would like to have my solution to IMO 95 A2 checked. All solutions I've found either used Cauchy-Schwarz, Chebyshevs inequality, the rearrangement inequality or Muirheads inequality. Me myself, I've used Jensens inequality, and I am a bit unsure if my solution holds. Here it is:
Let $a,b,c$ be positive real numbers with $abc=1$. Show that $$frac{1}{a^3(b+c)} + frac{1}{b^3(a+c)} + frac{1}{c^3(a+b)}geq frac32$$
First, substitute $frac1a = x, frac1b = y, frac1c = z$, giving the new constraint $xyz=1$, and transforming the inequality into $$frac{x^2}{z+y} + frac{y^2}{x+z} + frac{z^2}{x+y} geq frac32$$ Now, let $f(x)=dfrac{x^2}{S-x}$ where $S=x+y+z$. We have that $f''(x)=dfrac{2S^2}{(S-x)^2}$, and $S>0$ beacuse $x,y,z>0$, thus $f''(x)>0$ for all $x,y,zin mathbb{R}^+$. Thus we see that $f(x)$ is convex, so by Jensens inequality we have that $$begin{alignat*}{2}frac{x^2}{y+z} + frac{y^2}{x+z} + frac{z^2}{x+y} & =frac{x^2}{S-x} + frac{y^2}{S-y} + frac{z^2}{S-z} \
&geq 3fleft(frac{x+y+z}{3} right) \
&= 3frac{left(frac{x+y+z}{3}right)^2}{S-frac{x+y+z}{3}} = frac{1}{3}frac{(x+y+z)^2}{frac{2x+2y+2z}{3}} = frac13 frac{(x+y+z)^2}{frac23 (x+y+z)} \
&= frac12 (x+y+z) geq frac32end{alignat*}$$
The final inequality follows by AM-GM ($x+y+zgeq 3sqrt[3]{xyz}=3$)
proof-verification inequality contest-math alternative-proof jensen-inequality
edited Dec 16 '18 at 20:07
Blue
49k870156
asked Dec 16 '18 at 19:49
MarkusMarkus
9119
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add a comment |
$begingroup$
I would like to have my solution to IMO 95 A2 checked. All solutions I've found either used Cauchy-Schwarz, Chebyshevs inequality, the rearrangement inequality or Muirheads inequality. Me myself, I've used Jensens inequality, and I am a bit unsure if my solution holds. Here it is:
Let $a,b,c$ be positive real numbers with $abc=1$. Show that $$frac{1}{a^3(b+c)} + frac{1}{b^3(a+c)} + frac{1}{c^3(a+b)}geq frac32$$
First, substitute $frac1a = x, frac1b = y, frac1c = z$, giving the new constraint $xyz=1$, and transforming the inequality into $$frac{x^2}{z+y} + frac{y^2}{x+z} + frac{z^2}{x+y} geq frac32$$ Now, let $f(x)=dfrac{x^2}{S-x}$ where $S=x+y+z$. We have that $f''(x)=dfrac{2S^2}{(S-x)^2}$, and $S>0$ beacuse $x,y,z>0$, thus $f''(x)>0$ for all $x,y,zin mathbb{R}^+$. Thus we see that $f(x)$ is convex, so by Jensens inequality we have that $$begin{alignat*}{2}frac{x^2}{y+z} + frac{y^2}{x+z} + frac{z^2}{x+y} & =frac{x^2}{S-x} + frac{y^2}{S-y} + frac{z^2}{S-z} \
&geq 3fleft(frac{x+y+z}{3} right) \
&= 3frac{left(frac{x+y+z}{3}right)^2}{S-frac{x+y+z}{3}} = frac{1}{3}frac{(x+y+z)^2}{frac{2x+2y+2z}{3}} = frac13 frac{(x+y+z)^2}{frac23 (x+y+z)} \
&= frac12 (x+y+z) geq frac32end{alignat*}$$
The final inequality follows by AM-GM ($x+y+zgeq 3sqrt[3]{xyz}=3$)
proof-verification inequality contest-math alternative-proof jensen-inequality
edited Dec 16 '18 at 20:07
Blue
49k870156
asked Dec 16 '18 at 19:49
MarkusMarkus
9119
$endgroup$
add a comment |
$begingroup$
I would like to have my solution to IMO 95 A2 checked. All solutions I've found either used Cauchy-Schwarz, Chebyshevs inequality, the rearrangement inequality or Muirheads inequality. Me myself, I've used Jensens inequality, and I am a bit unsure if my solution holds. Here it is:
Let $a,b,c$ be positive real numbers with $abc=1$. Show that $$frac{1}{a^3(b+c)} + frac{1}{b^3(a+c)} + frac{1}{c^3(a+b)}geq frac32$$
First, substitute $frac1a = x, frac1b = y, frac1c = z$, giving the new constraint $xyz=1$, and transforming the inequality into $$frac{x^2}{z+y} + frac{y^2}{x+z} + frac{z^2}{x+y} geq frac32$$ Now, let $f(x)=dfrac{x^2}{S-x}$ where $S=x+y+z$. We have that $f''(x)=dfrac{2S^2}{(S-x)^2}$, and $S>0$ beacuse $x,y,z>0$, thus $f''(x)>0$ for all $x,y,zin mathbb{R}^+$. Thus we see that $f(x)$ is convex, so by Jensens inequality we have that $$begin{alignat*}{2}frac{x^2}{y+z} + frac{y^2}{x+z} + frac{z^2}{x+y} & =frac{x^2}{S-x} + frac{y^2}{S-y} + frac{z^2}{S-z} \
&geq 3fleft(frac{x+y+z}{3} right) \
&= 3frac{left(frac{x+y+z}{3}right)^2}{S-frac{x+y+z}{3}} = frac{1}{3}frac{(x+y+z)^2}{frac{2x+2y+2z}{3}} = frac13 frac{(x+y+z)^2}{frac23 (x+y+z)} \
&= frac12 (x+y+z) geq frac32end{alignat*}$$
The final inequality follows by AM-GM ($x+y+zgeq 3sqrt[3]{xyz}=3$)
proof-verification inequality contest-math alternative-proof jensen-inequality
edited Dec 16 '18 at 20:07
Blue
49k870156
asked Dec 16 '18 at 19:49
MarkusMarkus
9119
$endgroup$
I would like to have my solution to IMO 95 A2 checked. All solutions I've found either used Cauchy-Schwarz, Chebyshevs inequality, the rearrangement inequality or Muirheads inequality. Me myself, I've used Jensens inequality, and I am a bit unsure if my solution holds. Here it is:
Let $a,b,c$ be positive real numbers with $abc=1$. Show that $$frac{1}{a^3(b+c)} + frac{1}{b^3(a+c)} + frac{1}{c^3(a+b)}geq frac32$$
First, substitute $frac1a = x, frac1b = y, frac1c = z$, giving the new constraint $xyz=1$, and transforming the inequality into $$frac{x^2}{z+y} + frac{y^2}{x+z} + frac{z^2}{x+y} geq frac32$$ Now, let $f(x)=dfrac{x^2}{S-x}$ where $S=x+y+z$. We have that $f''(x)=dfrac{2S^2}{(S-x)^2}$, and $S>0$ beacuse $x,y,z>0$, thus $f''(x)>0$ for all $x,y,zin mathbb{R}^+$. Thus we see that $f(x)$ is convex, so by Jensens inequality we have that $$begin{alignat*}{2}frac{x^2}{y+z} + frac{y^2}{x+z} + frac{z^2}{x+y} & =frac{x^2}{S-x} + frac{y^2}{S-y} + frac{z^2}{S-z} \
&geq 3fleft(frac{x+y+z}{3} right) \
&= 3frac{left(frac{x+y+z}{3}right)^2}{S-frac{x+y+z}{3}} = frac{1}{3}frac{(x+y+z)^2}{frac{2x+2y+2z}{3}} = frac13 frac{(x+y+z)^2}{frac23 (x+y+z)} \
&= frac12 (x+y+z) geq frac32end{alignat*}$$
The final inequality follows by AM-GM ($x+y+zgeq 3sqrt[3]{xyz}=3$)
proof-verification inequality contest-math alternative-proof jensen-inequality
proof-verification inequality contest-math alternative-proof jensen-inequality
edited Dec 16 '18 at 20:07
Blue
49k870156
asked Dec 16 '18 at 19:49
MarkusMarkus
9119
edited Dec 16 '18 at 20:07
Blue
49k870156
asked Dec 16 '18 at 19:49
MarkusMarkus
9119
edited Dec 16 '18 at 20:07
Blue
49k870156
edited Dec 16 '18 at 20:07
Blue
49k870156
edited Dec 16 '18 at 20:07
Blue
49k870156
49k870156
asked Dec 16 '18 at 19:49
MarkusMarkus
9119
asked Dec 16 '18 at 19:49
MarkusMarkus
9119
asked Dec 16 '18 at 19:49
MarkusMarkus
9119
9119
add a comment |
add a comment |
3 Answers
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Your proof is nearly correct: $$f''(x)=frac{2S^2}{(S-x)^3}$$ (the exponent is $3$, not $2$). This is still okay because $x,y,z>0$, but you need a bit more justification.
answered Dec 16 '18 at 19:59
Carl SchildkrautCarl Schildkraut
11.7k11443
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You make a little mistake i think: $$f''(x)=frac{2x^2}{(S-x)^3}$$ (and the exponent is $3$, not $2$ in a numerator). Apart from that it is ok.
answered Dec 16 '18 at 20:02
Maria MazurMaria Mazur
46.4k1260119
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I think it's better to end your proof by C-S and AM-GM:
$$sum_{cyc}frac{x^2}{y+z}geqfrac{(x+y+z)^2}{sumlimits_{cyc}(y+z)}=frac{1}{2}(x+y+z)geqfrac{3}{2}.$$
answered Dec 16 '18 at 22:02
Michael RozenbergMichael Rozenberg
107k1895199
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
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votes
$begingroup$
Your proof is nearly correct: $$f''(x)=frac{2S^2}{(S-x)^3}$$ (the exponent is $3$, not $2$). This is still okay because $x,y,z>0$, but you need a bit more justification.
answered Dec 16 '18 at 19:59
Carl SchildkrautCarl Schildkraut
11.7k11443
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add a comment |
$begingroup$
Your proof is nearly correct: $$f''(x)=frac{2S^2}{(S-x)^3}$$ (the exponent is $3$, not $2$). This is still okay because $x,y,z>0$, but you need a bit more justification.
answered Dec 16 '18 at 19:59
Carl SchildkrautCarl Schildkraut
11.7k11443
$endgroup$
add a comment |
$begingroup$
Your proof is nearly correct: $$f''(x)=frac{2S^2}{(S-x)^3}$$ (the exponent is $3$, not $2$). This is still okay because $x,y,z>0$, but you need a bit more justification.
answered Dec 16 '18 at 19:59
Carl SchildkrautCarl Schildkraut
11.7k11443
$endgroup$
Your proof is nearly correct: $$f''(x)=frac{2S^2}{(S-x)^3}$$ (the exponent is $3$, not $2$). This is still okay because $x,y,z>0$, but you need a bit more justification.
answered Dec 16 '18 at 19:59
Carl SchildkrautCarl Schildkraut
11.7k11443
answered Dec 16 '18 at 19:59
Carl SchildkrautCarl Schildkraut
11.7k11443
answered Dec 16 '18 at 19:59
Carl SchildkrautCarl Schildkraut
11.7k11443
answered Dec 16 '18 at 19:59
Carl SchildkrautCarl Schildkraut
11.7k11443
11.7k11443
add a comment |
add a comment |
$begingroup$
You make a little mistake i think: $$f''(x)=frac{2x^2}{(S-x)^3}$$ (and the exponent is $3$, not $2$ in a numerator). Apart from that it is ok.
answered Dec 16 '18 at 20:02
Maria MazurMaria Mazur
46.4k1260119
$endgroup$
add a comment |
$begingroup$
You make a little mistake i think: $$f''(x)=frac{2x^2}{(S-x)^3}$$ (and the exponent is $3$, not $2$ in a numerator). Apart from that it is ok.
answered Dec 16 '18 at 20:02
Maria MazurMaria Mazur
46.4k1260119
$endgroup$
add a comment |
$begingroup$
You make a little mistake i think: $$f''(x)=frac{2x^2}{(S-x)^3}$$ (and the exponent is $3$, not $2$ in a numerator). Apart from that it is ok.
answered Dec 16 '18 at 20:02
Maria MazurMaria Mazur
46.4k1260119
$endgroup$
You make a little mistake i think: $$f''(x)=frac{2x^2}{(S-x)^3}$$ (and the exponent is $3$, not $2$ in a numerator). Apart from that it is ok.
answered Dec 16 '18 at 20:02
Maria MazurMaria Mazur
46.4k1260119
answered Dec 16 '18 at 20:02
Maria MazurMaria Mazur
46.4k1260119
answered Dec 16 '18 at 20:02
Maria MazurMaria Mazur
46.4k1260119
answered Dec 16 '18 at 20:02
Maria MazurMaria Mazur
46.4k1260119
46.4k1260119
add a comment |
add a comment |
$begingroup$
I think it's better to end your proof by C-S and AM-GM:
$$sum_{cyc}frac{x^2}{y+z}geqfrac{(x+y+z)^2}{sumlimits_{cyc}(y+z)}=frac{1}{2}(x+y+z)geqfrac{3}{2}.$$
answered Dec 16 '18 at 22:02
Michael RozenbergMichael Rozenberg
107k1895199
$endgroup$
add a comment |
$begingroup$
I think it's better to end your proof by C-S and AM-GM:
$$sum_{cyc}frac{x^2}{y+z}geqfrac{(x+y+z)^2}{sumlimits_{cyc}(y+z)}=frac{1}{2}(x+y+z)geqfrac{3}{2}.$$
answered Dec 16 '18 at 22:02
Michael RozenbergMichael Rozenberg
107k1895199
$endgroup$
add a comment |
$begingroup$
I think it's better to end your proof by C-S and AM-GM:
$$sum_{cyc}frac{x^2}{y+z}geqfrac{(x+y+z)^2}{sumlimits_{cyc}(y+z)}=frac{1}{2}(x+y+z)geqfrac{3}{2}.$$
answered Dec 16 '18 at 22:02
Michael RozenbergMichael Rozenberg
107k1895199
$endgroup$
I think it's better to end your proof by C-S and AM-GM:
$$sum_{cyc}frac{x^2}{y+z}geqfrac{(x+y+z)^2}{sumlimits_{cyc}(y+z)}=frac{1}{2}(x+y+z)geqfrac{3}{2}.$$
answered Dec 16 '18 at 22:02
Michael RozenbergMichael Rozenberg
107k1895199
answered Dec 16 '18 at 22:02
Michael RozenbergMichael Rozenberg
107k1895199
answered Dec 16 '18 at 22:02
Michael RozenbergMichael Rozenberg
107k1895199
answered Dec 16 '18 at 22:02
Michael RozenbergMichael Rozenberg
107k1895199
107k1895199
add a comment |
add a comment |
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