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If $a$ and $b$ are positive real numbers such that $a+b=1$, then prove that $$\bigg(a+ \dfrac {1}{a}\bigg)^2 +\bigg(b+ \dfrac {1}{b}\bigg)^2 \ge \dfrac {25}{2}$$

My tries: I am really unable to see through it. I have solved many inequality problems but somehow I am unable to solve this. I do not think my working will be of any help, so I am not typing those expressions.

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marked as duplicate by Martin R, Namaste algebra-precalculus Sep 22 '17 at 0:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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By Cauchy we have

$$\bigg(a+ \dfrac {1}{a}\bigg)^2 +\bigg(b+ \dfrac {1}{b}\bigg)^2 \ge \dfrac {1}{2}(a+{1\over a}+b+{1\over b})^2$$

so we have to check if

$$1+{1\over a}+{1\over b} \geq 5$$

but this is true since $$a+b = (a+b)^2\geq 4ab$$

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  • $\begingroup$ Thank you very much for giving a solution using the great Cauchy Schwarz inequality!☺️☺️ $\endgroup$ – ami_ba Sep 21 '17 at 19:32
  • $\begingroup$ You are wellcome! $\endgroup$ – Maria Mazur Sep 21 '17 at 19:33
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$f(x) = \left( x + \frac{1}{x}\right)^2$ is a convex function and so has exactly one minimum value on any interval. By the same token, the function $g(a, b) = f(a) + f(b)$ is convex and has exactly one minimum value on any line segment. On the line $L_\epsilon = \{(a, b) | a > \epsilon, b > \epsilon, a + b = 1\}$ for $\epsilon$ arbitrarily small, the minimum of $g$ can't be at an endpoint, because $f$ diverges at zero. So there must be a minimum in the interior of $L_\epsilon$. If the minimum were anywhere but $a = b = \frac{1}{2}$, then swapping $b$ and $a$ would give you another minimum, contradicting convexity, so the minimum is $g\left( \frac{1}{2}, \frac{1}{2} \right) = \frac{25}{2}$.

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If you write $$ f(x)=\left(x+\frac1x\right)^2+\left(1-x+\frac1{1-x}\right)^2, $$ and look for the critical points (differentiate and equate to zero) for $0<x<1$, you will find that there is a minimum at $x=1/2$. So $$ f(x)\geq f(1/2)=\frac{25}2. $$

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