<s>[INST] question from South Korean selection exam 1998, about proving that an inequality holds true if $a+b+c=abc$ 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?
 [/INST] 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:
$$\sum_{cyc}\sqrt{\frac{bc}{(a+b)(a+c)}}\leq\frac{1}{2}\sum_{cyc}\left(\frac{c}{a+c}+\frac{b}{a+b}\right)=$$
$$=\frac{1}{2}\sum_{cyc}\left(\frac{c}{a+c}+\frac{a}{c+a}\right)=\frac{3}{2}.$$
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