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.
$f(x) = \left( x + \frac{1}{x}\right)^2$ 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 $g\left( \frac{1}{2}, \frac{1}{2} \right) = \frac{25}{2}$.
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 $$ f(x)\geq f(1/2)=\frac{25}2. $$
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 $$a+b = (a+b)^2\geq 4ab$$
