<s>[INST] Prove $\frac{a^2}{3^3}+\frac{b^2}{4^3}+\frac{c^2}{5^3} \ge \frac{(a+b+c)^2}{6^3}$ If $a,b,c \in \mathbb{R+, }$ Then Prove that $$\frac{a^2}{3^3}+\frac{b^2}{4^3}+\frac{c^2}{5^3} \ge \frac{(a+b+c)^2}{6^3}$$
My try:
Consider $$P=\frac{a}{3\sqrt{3}}+\frac{b}{4\sqrt{4}}+\frac{c}{5\sqrt{5}}$$
BY Cauchy Scwartz Inequality we have
$$P \le \sqrt{3} \times \sqrt {\left(\frac{a^2}{3^3}+\frac{b^2}{4^3}+\frac{c^2}{5^3} \right)}$$
any way proceed here?
 [/INST] Using Titu's lemma(special case of CS inequiality): $$\frac{a^2}{3^3}+\frac{b^2}{4^3}+\frac{c^2}{5^3} \ge \frac{(a+b+c)^2}{(3^3+4^3+5^3)}=\frac{(a+b+c)^2}{6^3} $$
</s>