question from South Korean selection exam 1998, about proving that an inequality holds true if $a+b+c=abc$

Question

I was just doing the following question:

If $a,b,c>0$ such that $a+b+c=abc$, 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 think that this question can be solved through the use of homogenization, something which I attempted to do in the following way:

We have that $\frac{a+b+c}{abc}=1$ and hence also $\sqrt{\frac{a+b+c}{abc}}=1$. So $\frac{3}{2}*\sqrt{\frac{a+b+c}{abc}}=\frac{3}{2}$.

So all we have to prove now is $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}\le\frac{3}{2}*\sqrt{\frac{a+b+c}{abc}}$ which are homogenized and hence we do not need the original equality any more. This is where I couldn't continue from, and got stuck. Could you please explain to me how I could finish it off like this, or tell me why it can't and how it can be done using homogenization?

Answer 1

The homogenization gives: $$\sum_{cyc}\frac{1}{\sqrt{\frac{abc}{a+b+c}+a^2}}\leq\frac{3}{2}\sqrt{\frac{a+b+c}{abc}}$$ or $$\sum_{cyc}\sqrt{\frac{bc}{(a+b)(a+c)}}\leq\frac{3}{2},$$ which is true by AM-GM: $$\sum_{cyc}\sqrt{\frac{bc}{(a+b)(a+c)}}\leq\frac{1}{2}\sum_{cyc}\left(\frac{c}{a+c}+\frac{b}{a+b}\right)=$$ $$=\frac{1}{2}\sum_{cyc}\left(\frac{c}{a+c}+\frac{a}{c+a}\right)=\frac{3}{2}.$$

Answer 2

A different approach, but the desired result is equivalent to $\cos A+\cos B+\cos C\le\tfrac32$ in acute-angled triangles, which is true by Jensen's inequality.

Answer 3

Multiply the first term top & bottom by $\sqrt{bc}$ & use the constraint \begin{eqnarray*} \frac{\sqrt{bc}}{\sqrt{(a+b)(a+c)}}. \end{eqnarray*} Now use AM-GM \begin{eqnarray*} \frac{\sqrt{bc}}{\sqrt{(a+b)(a+c)}}<\frac{1}{2}\left( \frac{b}{a+b} +\frac{c}{a+c} \right). \end{eqnarray*}

Answer 4

We write inequality as $$\sum \sqrt{\frac{bc}{(a+b)(a+c)}} \leqslant \frac{3}{2}.$$ Using known inequality $$(a+b)(b+c)(c+a) \geqslant \frac{8}{9}(a+b+c)(ab+bc+ca),$$ and the Cauchy-Schwarz inequality we have $$\sum \sqrt{\frac{bc}{(a+b)(a+c)}} \leqslant \sqrt{(ab+bc+ca) \sum \frac{1}{(a+b)(a+c)}} $$ $$= \sqrt{\frac{2(ab+bc+ca)(a+b+c)}{(a+b)(b+c)(c+a)}} \leqslant \frac 32.$$ Done.