Understanding the proof of an inequality
Basically the method applied is the following, we fix $a=frac{a_1+a_2+...+a_n}{n}$, if: $$f(x)ge f(a)+f'(a)(x-a) $$This inequality holds for all x, then summing up the inequality will give us the desired conclusion.
That is basically what is stated before the example, now, I've got quite some questions!
1) Why do we have $a=frac{a_1+a_2+...+a_n}{n}$? why is this relevant?
2) When he applies this "tangent line trick" he clearly chooses $a=1$, is there a reason for this particular value? (I mean, there must be a reason since one has to bash through a lot of algebra, no one would do that if they didn't have a good reason to do so...)
3) The last inequality holds for $a ge- frac{1}{2}$, not all a's, so why does this still satisfy our initial method's constraint of 'all x'?
Edit: OP forgot to mention that variables are positive
inequality functional-inequalities tangent-line-method
edited Nov 28 '18 at 21:58
greedoid
38.2k114797
asked Nov 28 '18 at 21:17
Spasoje Durovic
34110
add a comment |
Basically the method applied is the following, we fix $a=frac{a_1+a_2+...+a_n}{n}$, if: $$f(x)ge f(a)+f'(a)(x-a) $$This inequality holds for all x, then summing up the inequality will give us the desired conclusion.
That is basically what is stated before the example, now, I've got quite some questions!
1) Why do we have $a=frac{a_1+a_2+...+a_n}{n}$? why is this relevant?
2) When he applies this "tangent line trick" he clearly chooses $a=1$, is there a reason for this particular value? (I mean, there must be a reason since one has to bash through a lot of algebra, no one would do that if they didn't have a good reason to do so...)
3) The last inequality holds for $a ge- frac{1}{2}$, not all a's, so why does this still satisfy our initial method's constraint of 'all x'?
Edit: OP forgot to mention that variables are positive
inequality functional-inequalities tangent-line-method
edited Nov 28 '18 at 21:58
greedoid
38.2k114797
asked Nov 28 '18 at 21:17
Spasoje Durovic
34110
add a comment |
Basically the method applied is the following, we fix $a=frac{a_1+a_2+...+a_n}{n}$, if: $$f(x)ge f(a)+f'(a)(x-a) $$This inequality holds for all x, then summing up the inequality will give us the desired conclusion.
That is basically what is stated before the example, now, I've got quite some questions!
1) Why do we have $a=frac{a_1+a_2+...+a_n}{n}$? why is this relevant?
2) When he applies this "tangent line trick" he clearly chooses $a=1$, is there a reason for this particular value? (I mean, there must be a reason since one has to bash through a lot of algebra, no one would do that if they didn't have a good reason to do so...)
3) The last inequality holds for $a ge- frac{1}{2}$, not all a's, so why does this still satisfy our initial method's constraint of 'all x'?
Edit: OP forgot to mention that variables are positive
inequality functional-inequalities tangent-line-method
edited Nov 28 '18 at 21:58
greedoid
38.2k114797
asked Nov 28 '18 at 21:17
Spasoje Durovic
34110
Basically the method applied is the following, we fix $a=frac{a_1+a_2+...+a_n}{n}$, if: $$f(x)ge f(a)+f'(a)(x-a) $$This inequality holds for all x, then summing up the inequality will give us the desired conclusion.
That is basically what is stated before the example, now, I've got quite some questions!
1) Why do we have $a=frac{a_1+a_2+...+a_n}{n}$? why is this relevant?
2) When he applies this "tangent line trick" he clearly chooses $a=1$, is there a reason for this particular value? (I mean, there must be a reason since one has to bash through a lot of algebra, no one would do that if they didn't have a good reason to do so...)
3) The last inequality holds for $a ge- frac{1}{2}$, not all a's, so why does this still satisfy our initial method's constraint of 'all x'?
Edit: OP forgot to mention that variables are positive
inequality functional-inequalities tangent-line-method
inequality functional-inequalities tangent-line-method
edited Nov 28 '18 at 21:58
greedoid
38.2k114797
asked Nov 28 '18 at 21:17
Spasoje Durovic
34110
edited Nov 28 '18 at 21:58
greedoid
38.2k114797
asked Nov 28 '18 at 21:17
Spasoje Durovic
34110
edited Nov 28 '18 at 21:58
greedoid
38.2k114797
edited Nov 28 '18 at 21:58
greedoid
38.2k114797
edited Nov 28 '18 at 21:58
greedoid
38.2k114797
38.2k114797
asked Nov 28 '18 at 21:17
Spasoje Durovic
34110
asked Nov 28 '18 at 21:17
Spasoje Durovic
34110
asked Nov 28 '18 at 21:17
Spasoje Durovic
34110
34110
add a comment |
add a comment |
1 Answer
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We see that equality is if $a=b=c=1$, so write a tangent on $$f(x) = {(3-2x)^2over x^2+(3-x)^2}$$ at $x_0=1$ and $y_0= f(1) = 1/5$. This line has equation $$y = {1over 5}- {18over 25}(x-1)$$
Now check if $f(x)geq y$ for all $xin (0,3)$ and thus a conclusion.
answered Nov 28 '18 at 21:25
greedoid
38.2k114797
add a comment |
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1 Answer
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1 Answer
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We see that equality is if $a=b=c=1$, so write a tangent on $$f(x) = {(3-2x)^2over x^2+(3-x)^2}$$ at $x_0=1$ and $y_0= f(1) = 1/5$. This line has equation $$y = {1over 5}- {18over 25}(x-1)$$
Now check if $f(x)geq y$ for all $xin (0,3)$ and thus a conclusion.
answered Nov 28 '18 at 21:25
greedoid
38.2k114797
add a comment |
We see that equality is if $a=b=c=1$, so write a tangent on $$f(x) = {(3-2x)^2over x^2+(3-x)^2}$$ at $x_0=1$ and $y_0= f(1) = 1/5$. This line has equation $$y = {1over 5}- {18over 25}(x-1)$$
Now check if $f(x)geq y$ for all $xin (0,3)$ and thus a conclusion.
answered Nov 28 '18 at 21:25
greedoid
38.2k114797
add a comment |
We see that equality is if $a=b=c=1$, so write a tangent on $$f(x) = {(3-2x)^2over x^2+(3-x)^2}$$ at $x_0=1$ and $y_0= f(1) = 1/5$. This line has equation $$y = {1over 5}- {18over 25}(x-1)$$
Now check if $f(x)geq y$ for all $xin (0,3)$ and thus a conclusion.
answered Nov 28 '18 at 21:25
greedoid
38.2k114797
We see that equality is if $a=b=c=1$, so write a tangent on $$f(x) = {(3-2x)^2over x^2+(3-x)^2}$$ at $x_0=1$ and $y_0= f(1) = 1/5$. This line has equation $$y = {1over 5}- {18over 25}(x-1)$$
Now check if $f(x)geq y$ for all $xin (0,3)$ and thus a conclusion.
answered Nov 28 '18 at 21:25
greedoid
38.2k114797
answered Nov 28 '18 at 21:25
greedoid
38.2k114797
answered Nov 28 '18 at 21:25
greedoid
38.2k114797
answered Nov 28 '18 at 21:25
greedoid
38.2k114797
38.2k114797
add a comment |
add a comment |
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