Let $a;b;c\in R^+$ such that $ab+bc+ca>0$. Prove that $$\sqrt{\frac{a^2+1}{b+c}}+\sqrt{\frac{b^2+1}{a+c}}+\sqrt{\frac{c^2+1}{a+b}}\ge 3$$

I have seen the similar question is $$\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b}\ge 3$$ my problem is harder than it

My try: Holder: $$LHS^2\cdot \sum _{cyc}\left(a^2+1\right)^2\left(b+c\right)\ge \left(a^2+b^2+c^2+3\right)^3$$

Then we need to prove: $$\left(a^2+b^2+c^2+3\right)^3\ge 9\sum _{cyc}\left(a^2+1\right)^2\left(b+c\right)$$

After full expanding I am stuck because that inequality is not homogeneous, it hard to solve.


1 Answer 1


Because by AM-GM and C-S we obtain: $$\sum_{cyc}\sqrt{\frac{a^2+1}{b+c}}\geq3\sqrt[6]{\prod_{cyc}\frac{a^2+1}{b+c}}=3\sqrt[12]{\prod_{cyc}\frac{(a^2+1)(1+b^2)}{(a+b)^2}}\geq3\sqrt[12]{\prod_{cyc}\frac{(a+b)^2}{(a+b)^2}}=3$$


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.