Show that $(a+2b+\frac{2}{a+1})(b+2a+\frac{2}{b+1})\geqslant16$ if $ ab\geqslant1$

Question

$a$, $b$ are positive real numbers.

Show that $$\left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right)\geqslant16$$ if $$ ab\geqslant1$$

Should I use AM-GM inequality? If yes, where?

Answer 1

Let $a+b=2u$.

Thus, by AM-GM $u\geq1$ and by C-S and AM-GM we obtain: $$\left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right)=$$

$$=\left(a+b+b+\frac{2}{a+1}\right)\left(a+b+a+\frac{2}{b+1}\right)\geq$$ $$\geq\left(a+b+\sqrt{ab}+\frac{2}{\sqrt{(a+1)(b+1)}}\right)^2\geq$$ $$\geq\left(2u+1+\frac{4}{a+1+b+1}\right)^2=\left(\frac{u+1}{2}+\frac{2}{u+1}+\frac{3}{2}u+\frac{1}{2}\right)^2\geq$$ $$\geq\left(2+\frac{3}{2}+\frac{1}{2}\right)^2=16.$$ Done!

Answer 2

Expand to get:

$$ab + 2a^2 + \frac{2a}{b+1} + 2b^2 + 4ab + \frac{4b}{b+1} + \frac{2b}{a+1} + \frac{4a}{a+1} + \frac{4}{(a+1)(b+1)} \ge 9 + \frac{2a^2 + 2a + 4ab + 4b + 2b^2 + 2b + 4ab + 4a + 4}{(a+1)(b+1)} \ge 9 + 7 = 16$$

The last inequality follows as:

$$2a^2 + 2a + 4ab + 4b + 2b^2 + 2b + 4ab + 4a + 4 \ge 7(a+1)(b+1)$$ $$2a^2 + 2b^2 + ab \ge a+ b + 3$$

Now it's enough to prove that $2a^2 + 2b^2 \ge a + b + 2$. From the condition we have by AM-GM $a+b \ge 2\sqrt{ab} \ge 2$. Hence by using the quadratic mean inequality:

$$2a^2 + 2b^2 \ge (a+b)^2 \ge 2(a+b) \ge a+b+2$$

Hence the proof.