Solving an equality in 2 variables I need to prove that
$$\left(a + \frac{1}{a}\right)^2 +\left(b + \frac{1}{b}\right)^2 \gt \frac{25}{2}$$
if $a+b = 1$  and $a b \le 1/4$
I'd like a hint. Solve the equality first to $a$ or $b$, or stay in a and b as to get
$a b \le 4$ in the inequality ?

Apply Cauchy-Schwarz Inequality to argue that,
$(a+\frac{1}{a}+b+\frac{1}{b})^2 \le (1+1)((a+\frac{1}{a})^2+(b+\frac{1}{b})^2)$,
And further, $\frac{1}{a}+\frac{1}{b} = \frac{a+b}{ab}=\frac{1}{ab}\ge \frac{4}{(a+b)^2}=4$.
So the first expression becomes, 
$25=(1+4)^2 \le (a+\frac{1}{a}+b+\frac{1}{b})^2 \le (1+1)((a+\frac{1}{a})^2+(b+\frac{1}{b})^2)$
That is $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2 \ge 25/2$.
and Equality holds iff $a=b=\frac12$.
