How can I prove this inequality? $(a^2+1)(b^2+1)>a(b^2+1)+b(a^2+1)$

Question

Given that $a,b$ are real numbers. How can one show that $(a^2+1)(b^2+1)>a(b^2+1)+b(a^2+1)$ ?!!

Answer 1

Notice that, given a real number $x$,

$$(x-1)^2=x^2-2x+1\geq 0$$

Thus

$$2x\leq 1+x^2$$

$$\frac{2x}{1+x^2}\leq 1$$

Then, with $x=a$ then $x=b$,

$$\frac{2a}{1+a^2}+\frac{2b}{1+b^2}\leq 2$$

Divide by $2$

$$\frac{a}{1+a^2}+\frac{b}{1+b^2}\leq 1$$

Multiply by $(a^2+1)(b^2+1)$

$$a(b^2+1)+b(a^2+1)\leq (a^2+1)(b^2+1)$$

Notice that the inequality is not strict: when $a=b=1$, you have an equality. From the preceding derivation, it's clear that this is the only case of equality.

Answer 2

The inequality must not be a strict inequality. Rewrite it as: $$\Bigl(a+\frac1a\Bigr)\Bigl(b+\frac1b\Bigr)\ge\Bigl(a+\frac1a\Bigr)+\Bigl(b+\frac1b\Bigr)$$ and set $x=a+\dfrac1a, \enspace y=b+\dfrac1b$.

By the arithmetic means-geometric means inequality, one sees $x, y\ge 2$, so we have to prove: $$xy\ge x+y\quad\text{if}\quad x,y\ge 2.$$ This inequality can be rewritten as $\,\dfrac1x+\dfrac 1y\le 1$, which results from $\dfrac1x,\dfrac1y\le\dfrac12$.