Suppose $a,b,c\geq 0$ and $ab+bc+ca=1$, prove that $a\sqrt{a^2+1}+b\sqrt{b^2+1}+c\sqrt{c^2+1}\geq 2$.

Question

Suppose $a,b,c\geq 0$ and $ab+bc+ca=1$, prove that $$a\sqrt{a^2+1}+b\sqrt{b^2+1}+c\sqrt{c^2+1}\geq 2.$$ In my opinion, I do this problem by th following: $$\sum_{cyc}a\sqrt{a^2+1}=\sum_{cyc}a\sqrt{(a+b)(c+a)},$$ but I do not know how to continue following this way. It seems easy to enlarge this quantity, but how to reduce this quantity. Any help and hint will welcome!!

Answer 1

The case $abc = 0$ can be treated separately. Let's assume $abc > 0$, and put $a = \tan\left(\alpha/2\right), b = \tan\left(\beta/2\right), c = \tan\left(\gamma/2\right)\implies \alpha + \beta+\gamma = \pi$ . Hence we prove: $\displaystyle \Sigma \dfrac{\sin(\alpha/2)}{1-\sin^2(\alpha/2)}\ge 2$. Consider the function $f(x) = \dfrac{\sin x}{1-\sin^2 x}, 0 < x < \pi/2\implies f''(x) = \dfrac{-2+\sin x}{(1+\sin x)^3}+\dfrac{2+\sin x}{(1-\sin x)^3} > 0 $ ( can simplify this and see it positive ). Thus $f$ is convex, thus $\Sigma f(x)\ge 3f(\Sigma x/3)= 3f(\pi/6)=2$ as claimed.

Answer 2

I think your idea works, let's proceed:

Introduce the following substitution: $$a + b = u^2$$ $$b + c = v^2$$ $$c + a = w^2$$ Then we have: $$2a = u^2 + w^2 - v^2$$ $$2b = v^2 + u^2 - w^2$$ $$2c = w^2 + v^2 - u^2$$ $ab + bc + ca = 1$ becomes: $$4 = 2u^2v^2 + 2v^2w^2 + 2w^2u^2 - u^4 - v^4 - w^4$$ So the inequality becomes: $$\sum_{cyc}{(u^2 + w^2 - v^2)uw} \geq 2\sum_{cyc}{u^2v^2} - \sum_{cyc}u^4$$ In Muirhead-like notation: $$T[3, 1, 0] + \frac{1}{2}T[4, 0, 0] \geq T[2, 2, 0] + \frac{1}{2}T[2, 1, 1]$$ This is just pure Muirhead with $T[3, 1, 0] \geq T[2, 2, 0]$ and $T[4, 0, 0] \geq T[2,1,1]$.

Answer 3

By C-S, Schur and AM-GM we obtain:

$$\sum_{cyc}a\sqrt{a^2+1}=\sum_{cyc}a\sqrt{(a+b)(a+c)}\geq\sum_{cyc}a(a+\sqrt{bc})=$$

$$=\sum_{cyc}(a^2-\sqrt{a^3b}-\sqrt{a^3c}+a\sqrt{bc})+\sum_{cyc}(\sqrt{a^3b}+\sqrt{ab^3})\geq2\sum_{cyc}ab=2.$$