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

Let $a,b,$ and $c$ be positive real numbers such that $a+b+c = abc$. Prove that $$\dfrac{1}{\sqrt{1+a^2}}+\dfrac{1}{\sqrt{1+b^2}}+\dfrac{1}{\sqrt{1+c^2}} \leq \dfrac{3}{2}.$$

This question seems tricky because how do we incorporate the $a+b+c = abc$ condition to solve this? I was thinking of solving for the variables and substituting like $a = \dfrac{b+c}{1-bc}$ etc., but I am not sure how that will help.

Change the notation using $x=\frac{1}{a}$ , $y=\frac{1}{b}$ and $z=\frac{1}{c}$

Now the condition is that $xy+yz+zx=1$.

And the inequality to prove is :

$$\frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}} \leq \frac{3}{2}$$

Now use the condition and notice that :

$$x^2+1=x^2+xy+yz+zx=(x+y)(x+z)$$

So now using the AM-GM inequality :

$$\frac{x}{\sqrt{x^2+1}}=\frac{x}{\sqrt{(x+y)(x+z)}} =\sqrt{\frac{x^2}{(x+y)(x+z)}} \leq \frac{1}{2} \left (\frac{x}{x+y}+\frac{x}{x+z} \right )$$

Do the same thing for the other terms and add them to get the conclusion .

Let $a=\tan(A)$, $b=\tan(B)$ and $c=\tan(C)$. We then have $$\tan(A)+\tan(B)+\tan(C) = \tan(A)\tan(B)\tan(C) \,\,\,\, (\spadesuit)$$ Recall that $(\spadesuit)$ is true iff $A+B+C = \pi$, i.e., $A$, $B$ and $C$ are angles of a triangle. (Infact the part we need is trivial, since we have $\tan(C) = -\dfrac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)} = \tan(\pi-(A+B))$.)

Further, since $a,b,c > 0$, we have that $0 \leq A,B,C < \dfrac{\pi}2$. We then have that $$\dfrac1{\sqrt{1+a^2}} + \dfrac1{\sqrt{1+b^2}} + \dfrac1{{\sqrt{1+c^2}}} = \underbrace{\cos(A) + \cos(B) + \cos(C) \leq 3\cos\left (\dfrac{A+B+C}3\right)}_{\text{By Jensen}} = \dfrac32$$

• Nice technique ! – Gabriel Romon Jan 8 '16 at 18:22
• Why is $\spadesuit$ true iff $A+B+C = \pi$? – user19405892 Jan 8 '16 at 18:34
• @user19405892 I suggest that you take a look at the numerator of the tan(A+B+C) identity. – Nikunj Jan 8 '16 at 19:01