Prove that $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}\le\frac{3}{2}$

Question

For positive real numbers with $a+b+c=abc$ prove that

$$\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}\le\frac{3}{2}$$

I made the substitution $a=\tan(\alpha), b = \tan(\beta), c= \tan(\gamma)$ with the constraint $\alpha+\beta+\gamma=\pi$

My inequality reduces to proving,

$$\cos^2(\alpha)+\cos^2(\beta)+\cos^2(\gamma) \le\frac32$$

But I am stuck on it. Any help with either inequality would be appreciated.

Answer 1

Nope, reduces to $\sum \cos$ not squared. Then just a standard Jensen's argument. ;)

Answer 2

By AM-GM $$\sum_{cyc}\frac{1}{\sqrt{1+a^2}}=\sum_{cyc}\frac{1}{\sqrt{\frac{abc}{a+b+c}+a^2}}=\sum_{cyc}\sqrt{\frac{a+b+c}{a(a+b)(a+c)}}=$$ $$=\sum_{cyc}\sqrt{\frac{bc}{(a+b)(a+c)}}\leq\frac{1}{2}\sum_{cyc}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{b}{a+b}+\frac{a}{a+b}\right)=\frac{3}{2}.$$ Done!