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 ?
 A: 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 $a=b=\frac12$.
A: Hint:
Put $\;b=1-a\;$ in the left side of the inequality to prove, so:
$$\left(a+\frac1a\right)^2+\left(1-a+\frac1{1-a}\right)^2=a^2+2+\frac1{a^2}+(1-a)^2+2+\frac1{(1-a)^2}=$$
$$=2a^2-2a+5+\frac{2a^2-2a+1}{\left(a(1-a)\right)^2}$$
But it is given that
$$\frac14>ab=a(1-a)=a-a^2\iff a(a-1)=a^2-a>-\frac14\;\ldots$$
Try to take it from here now...and check the inequality sign.
A: $$
\begin{align}
\hspace{-1cm}\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2
&\ge2\left(a+\frac1a\right)\left(b+\frac1b\right)\tag{1}\\
&=\frac2{ab}\left(a^2+1\right)\left(b^2+1\right)\\
&\ge8\left(a^2+1\right)\left(b^2+1\right)\tag{2}\\
&=8\small\left(\left(a-\tfrac12\right)^2+\left(a-\tfrac12\right)+\tfrac54\right)\left(\left(a-\tfrac12\right)^2-\left(a-\tfrac12\right)+\tfrac54\right)\\
&=8\left[\left(\left(a-\tfrac12\right)^2+\tfrac54\right)^2-\left(a-\tfrac12\right)^2\right]\\
&=8\left[\left(a-\tfrac12\right)^4+\tfrac32\left(a-\tfrac12\right)^2+\tfrac{25}{16}\right]\\[4pt]
&\ge\frac{25}{2}\tag{3}
\end{align}
$$
Inequalities:
$(1)$: $(x-y)^2\ge0\implies x^2+y^2\ge2xy$
$(2)$: $ab\le\tfrac14$
$(3)$: $x^2\ge0$
