Prove $\frac{a^2}{3^3}+\frac{b^2}{4^3}+\frac{c^2}{5^3} \ge \frac{(a+b+c)^2}{6^3}$

Question

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?

Answer 1

Using Titu's lemma(special case of CS inequiality): $$\frac{a^2}{3^3}+\frac{b^2}{4^3}+\frac{c^2}{5^3} \ge \frac{(a+b+c)^2}{(3^3+4^3+5^3)}=\frac{(a+b+c)^2}{6^3} $$

Answer 2

For $\lambda = a+b+c$, Cauchy-Schwarz inequality gives $$\lambda^2 \leqslant (3^3 + 4^3+ 5^3)(\frac{a^2}{3^3} + \frac{b^2}{4^3} + \frac{c^2}{5^3})$$ and you can check that $3^3+4^3+5^3=6^3$.