Saudi Arabian Mathematical Competitions

First Solution. If $a = b$, the inequality is satisfied. If $a \neq b$, we can assume without loss of generality that $a > b$. In this case, the inequality can be written
$$\left(2^{n}-1\right)\left(a^{2^{n}}-b^{2^{n}}\right) \geq 2^{n} \sqrt{a b}\left(a^{2^{n}-1}-b^{2^{n}-1}\right),$$
or equivalently
$$2^{n}(\sqrt{a}-\sqrt{b})\left(a^{2^{n}-\frac{1}{2}}+b^{2^{n}-\frac{1}{2}}\right) \geq a^{2^{n}}-b^{2^{n}} .$$
By dividing in both sides by $\sqrt{a}-\sqrt{b}$, the inequality becomes
$$2^{n}\left(A^{2^{n+1}-1}+B^{2^{n+1}-1}\right) \geq (A+B)\left(A^{2}+B^{2}\right)\left(A^{4}+B^{4}\right) \cdots\left(A^{2^{n}}+B^{2^{n}}\right),$$
where $A = \sqrt{a}$ and $B = \sqrt{b}$. We prove this last inequality by induction on $n \geq 1$.

For $n = 1$, we have
$$\begin{aligned} (A+B)\left(A^{2}+B^{2}\right) & = A^{3}+B^{3}+A^{2} B+A B^{2} \\ & \leq A^{3}+B^{3}+\frac{2 A^{3}+B^{3}}{3}+\frac{A^{3}+2 B^{3}}{3} \\ & \leq 2\left(A^{3}+B^{3}\right), \end{aligned}$$
by AM-GM inequality.

Assume the inequality true for $n$. We have
$$\begin{aligned} \left(A^{2^{n+1}-1}+B^{2^{n+1}-1}\right)\left(A^{2^{n+1}}+B^{2^{n+1}}\right) & \\ = & A^{2^{n+2}-1}+B^{2^{n+2}-1}+A^{2^{n+1}-1} B^{2^{n+1}}+A^{2^{n+1}} B^{2^{n+1}-1} \\ \leq & A^{2^{n+2}-1}+B^{2^{n+2}-1}+\frac{\left(2^{n+1}-1\right) A^{2^{n+2}-1}+2^{n+1} B^{2^{n+2}-1}}{2^{n+2}-1} \\ & +\frac{2^{n+1} A^{2^{n+2}-1}+\left(2^{n+1}-1\right) B^{2^{n+2}-1}}{2^{n+2}-1} \\ \leq 2\left(A^{2^{n+2}-1}+B^{2^{n+2}-1}\right) & \end{aligned}$$
by AM-GM inequality. We deduce the inequality for $n+1$.

Second Solution. Consider the polynomial
$$P(X) = \left(1-2^{-n}\right)\left(X^{2^{n+1}}-1\right) - X\left(X^{2^{n+1}-2}-1\right) .$$
Its derivative polynomial is
$$P^{\prime}(X) = \left(\left(2^{n+1}-2\right) X - \left(2^{n+1}-1\right)\right) X^{2^{n+1}-2} + 1 .$$
Therefore, for all $x \geq 1$, $P^{\prime}(x) \geq P^{\prime}(1) = 0$. Hence, the polynomial function $P$ is increasing on $[1, \infty[$.

Let $a, b$ be two positive real numbers with $a \geq b$. Because $\sqrt{\frac{a}{b}} \geq 1$, we have
$$\begin{aligned} b^{2^{n}} P\left(\sqrt{\frac{a}{b}}\right) & = \left(1-2^{-n}\right) b^{2^{n}}\left(\left(\frac{a}{b}\right)^{2^{n}}-1\right) - b^{2^{n}} \sqrt{\frac{a}{b}}\left(\left(\frac{a}{b}\right)^{2^{n}-1}-1\right) \\ & = \left(1-2^{-n}\right)\left(a^{2^{n}}-b^{2^{n}}\right) - \sqrt{a b}\left(a^{2^{n}-1}-b^{2^{n}-1}\right) \geq b^{2^{n}} P(1) = 0 \end{aligned}$$
which proves the inequality.