<s>[INST] cyclic three variable inequality 
Let $a,b,c$ be nonnegative real numbers and $a+b+c=3$. Prove the inequality
  $$
  \sqrt{24a^2b+25}+\sqrt{24b^2c+25}+\sqrt{24c^2a+25}\le 21 
$$

I have tried to find the solution using classical inequalities, but failed. Any idea?
 [/INST] By C-S
$$\left(\sum_{cyc}\sqrt{24a^2b+25}\right)^2\leq\sum_{cyc}(24a^2b+25)(a+3b+5c)\sum_{cyc}\frac{1}{a+3b+5c}$$
Thus, it remains to prove that
$$\sum_{cyc}(24a^2b+25)(a+3b+5c)\sum_{cyc}\frac{1}{a+3b+5c}\leq441$$ or
$$\sum_{cyc}(648a^2b+25(a+b+c)^3)(a+3b+5c)\sum_{cyc}\frac{1}{a+3b+5c}\leq441(a+b+c)^3$$ or
$$\sum_{cyc}(20a^6+47a^5b+352a^5c-209a^4b^2+509a^4c^2-275a^3b^3+$$
$$+690a^4bc-556a^3b^2c-143a^3c^2b-435a^2b^2c^2)\geq0.$$
Now, let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Hence, we need to prove that
$$4779(u^2-uv+v^2)a^4+27(263u^3-211u^2v+130uv^2+263v^3)a^3+$$
$$+9(365u^4-58u^3v-453u^2v^2+965uv^3+365v^4)a^2+$$
$$+3(173u^5+169u^4v-739u^3v^2+356u^2v^3+1156uv^4+173v^5)a+$$
$$+(u-2v)^2(20u^4+127u^3v+219u^2v^2+93uv^3+5v^4)\geq0,$$
which is obvious.
Done!
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