5

$a,b,c$ are real numbers $>0$. If $a+b+c=1$, show that

$$(a+\frac{1}{a})^2+(b+\frac{1}{b})^2+(c+\frac{1}{c})^2\geq\frac{100}{3}$$

choco_addicted
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Soham
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  • I have tried applying AM-GM separately on all terms... – Soham Feb 25 '16 at 15:55
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    Actually, the conclusion, as written, is not quite true. Consider the case where a= b= c= 1/3. Then the sum of squares is equal to 100/3, not greater. I presume you intended $\ge \frac{100}{3}$. – user247327 Feb 25 '16 at 15:57
  • Related : http://math.stackexchange.com/questions/487486/proving-inequality-a-frac1a2-b-frac1b2-geq-frac252-for – lab bhattacharjee Feb 25 '16 at 16:25

2 Answers2

9

The function $f(x) =(x+x^{-1} )^2 $ is convex on $(0,\infty) $ hence $$ \frac{1}{3} (f(a) +f(b) +f(c))\geq f(\frac{a+b+c}{3}) =(3+3^{-1} )^2 $$

4

Let $f(x)=(x+{1 \over x})^2$ and note that $f'$ is strictly increasing on $(0,\infty)$.

Let $\phi(x) = \{ f(x_1)+f(x_2)+f(x_3)$ and consider $\min \{ \phi(x) | \sum_k x_k = 1, x_i \ge 0 \}$. Since $\phi(x)$ is unbounded if any component of $x$ approaches zero, and the simplex is compact, we see that the problem has a solution $\hat{x}$ and $\hat{x}_k >0$ for all $k$. Hence we can apply Lagrange multipliers to get some $\lambda $ such that $f'(\hat{x}_k) + \lambda = 0$ for all $k$, and hence $f'(\hat{x}_1) = f'(\hat{x}_2) =f'(\hat{x}_3) $.

Since $f'$ is injective, we see that $\hat{x}_1 = \hat{x}_2 =\hat{x}_3 $, and hence $\hat{x}_1 = \hat{x}_2 =\hat{x}_3 = {1 \over 3}$ from which we get $\phi(x) \ge \phi(\hat{x}) = 3 f'({1 \over3}) = {100 \over 3}$.

copper.hat
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