Question

If $a$ and $b$ are positive and $a+b=1$, show that $(a+1/a{)}^2+(b+1/b{)}^2>\frac{25}{2}$.

Answer 1

$\Rightarrow$ a & b are the

$\Rightarrow a+b=1$

${(a+\frac{1}{a})}^2+{(b+\frac{1}{b})}^2>\frac{25}{2}$

L.H.S:-

$\Rightarrow 4+a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2}\ge 4+a^2+b^2+\frac{1}{a^2}+\frac{1}{b^2}$ ........$\left(1\right)$

We have canchy-schmorz

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

$(a+b)(a+b)(\frac{1}{a^2}+\frac{1}{b^2})\ge (1+1{)}^3$

from equation $\left(1\right)$

$\ge 4+\frac{1}{2}+8=\frac{25}{2}$ R.H.S.