63rd Czech and Slovak Mathematical Olympiad

It is evident that the inequality under consideration becomes an equality when $a=0$, $b=0$ or $a=b$. To prove that otherwise the strong inequality holds, it suffices to deal with the case $a>b>0$ and (after removing the fractions) to show that $$a\sqrt{a^2+1}+b\sqrt{b^2+1}>(a+b)\sqrt{a^2+1}$$ Distributing the right-hand side and regrouping the terms we get $$a\sqrt{a^2+1}(\sqrt{ab+1}-\sqrt{b^2+1})>b\sqrt{b^2+1}(\sqrt{a^2+1}-\sqrt{ab+1})$$ Multiplying the differences of the square roots by their sums as the denominators of new installed fractions, we obtain $$a\sqrt{a^2+1}\cdot \frac{b(a-b)}{\sqrt{ab+1}+\sqrt{b^2+1}}>b\sqrt{b^2+1}\cdot \frac{a(a-b)}{\sqrt{a^2+1}+\sqrt{ab+1}}$$ Dividing both sides by the positive number $ab(a-b)$ and removing the fractions again, we finally arrive at an equivalent inequality $$\sqrt{a^2+1}(\sqrt{a^2+1}+\sqrt{ab+1})>\sqrt{b^2+1}(\sqrt{b^2+1}+\sqrt{ab+1})$$ which easily follows by an easy comparison of the both sides "term by term" (because our assumption $a>b$ implies that $\sqrt{a^2+1}>\sqrt{b^2+1}$). This completes the proof of the given inequality. As we have shown, the only cases of the equality are $a=0$, $b=0$ and $a=b$.

Another solution: We exclude the cases $a=0$ and $b=0$ (when the inequality becomes an equality) from our considerations. Let us apply a Cauchy-Schwarz inequality in the form $$\left(\frac{a}{u}+\frac{b}{v}\right)(au+bv)\ge (a+b{)}^2,$$ with positive coefficients $u=\sqrt{b^2+1}$ and $v=\sqrt{a^2+1}$: $$\left(\frac{a}{\sqrt{b^2+1}}+\frac{b}{\sqrt{a^2+1}}\right)\left(a\sqrt{b^2+1}+b\sqrt{a^2+1}\right)\ge (a+b{)}^2.(1)$$ Another Cauchy-Schwarz inequality yields an upper bound for the second factor from the-left hand side of (1): $$\begin{array}{rl}a\sqrt{b^2+1}+b\sqrt{a^2+1}& =\sqrt{a}\sqrt{a{b}^2+a}+\sqrt{b}\sqrt{a^{2}b+b}\le \\ & \le \sqrt{a+b}\sqrt{a{b}^2+a+a^{2}b+b}=\sqrt{a+b}\sqrt{(a+b)(ab+1)}=(a+b)\sqrt{ab+1}.\end{array}$$ Consequently, the first factor in (1) has a lower bound $$\frac{a}{\sqrt{b^2+1}}+\frac{b}{\sqrt{a^2+1}}\ge \frac{(a+b{)}^2}{a\sqrt{b^2+1}+b\sqrt{a^2+1}}\ge \frac{a+b}{\sqrt{ab+1}},$$ which is the desired inequality. Since (1) becomes an equality if and only if the positive coefficients $u$ and $v$ are the same, i.e. $\sqrt{b^2+1}=\sqrt{a^2+1}$ in our situation, the equality $a=b$ is the third (and last) case (next to $a=0$ and $b=0$ from the introductory sentence) when the proven inequality holds as an equality.

Another solution: Let us exclude the obvious cases $a=0,b=0,a=b$ and let us transform the (strong) inequality under consideration into the following equivalent form: $$\frac{a}{a+b}\cdot \frac{1}{\sqrt{b^2+1}}+\frac{b}{a+b}\cdot \frac{1}{\sqrt{a^2+1}}>\frac{1}{\sqrt{ab+1}}(2)$$ The last left-hand side can be read as that of the (strong) Jensen inequality $$\mathrm{pf}(\alpha )+\mathrm{qf}(\beta )>f(\rho \alpha +q\beta ),(3)$$ with positive coefficients $p=a/(a+b)$ and $q=b/(a+b)$ (which satisfy $p+q=1$ as required), applied to the function $f(x)=1/\sqrt{x}$ at the points $\alpha =b^2+1$ and $\beta =a^2+1$. Since the function $f$ is strictly convex on the interval $(0,+\infty )$ and since the points $\alpha$ and $\beta$ are assumed to be distinct, the Jensen inequality (3) holds. It remains to verify that also the right-hand sides of (2) and (3) are identical. This is easy: $$\begin{array}{rl}f(p\alpha +q\beta )& =f\left(\frac{a}{a+b}(b^2+1)+\frac{b}{a+b}(a^2+1)\right)=\\ & =f\left(\frac{a+a{b}^2+b+a^{2}b}{a+b}\right)=f(ab+1)=\frac{1}{\sqrt{ab+1}}.\end{array}$$