2
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 .

$\endgroup$
5
$\begingroup$

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 $$\dfrac1{\sqrt{1+a^2}} + \dfrac1{\sqrt{1+b^2}} + \dfrac1{{\sqrt{1+c^2}}} = \underbrace{\cos(A) + \cos(B) + \cos(C) \leq 3\cos\left (\dfrac{A+B+C}3\right)}_{\text{By Jensen}} = \dfrac32$$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.