Use the pth power mean to prove an inequality

If $a^2 + b^2 + c^2 = 8$ for $a,b,c > 0$ , prove that

$a^4 + b^4 + c^4 \geq \frac{64}{3}.$

Proof. For $a,b,c \in \mathbb{R}$ not all equal, we may apply the previous exercise with $x_1 = a, \ x_2 = b, \ x_3 = c$ and with $p = 2$ , to obtain,

$\begin{align*} M_2 < M_4 \qquad \qquad &\implies& \left( \frac{a^2+b^2+c^}{3} \right)^{1/2} &< \left( \frac{a^4+b^4+c^4}{3} \right)^{1/4} \\ &\implies& \left( \frac{8}{3} \right)^2 &< \frac{a^4+b^4+c^4}{3} \\ &\implies& a^4 + b^4 + c^4 &> \frac{64}{3}. \end{align*}$

Further, if $a=b=c$ , then

$\begin{align*} a^2+b^2+c^2 = 8 &&\implies 3a^2 &= 8\\ &&\implies a^2 = b^2 = c^2 &= \frac{8}{3}\\ &&\implies a^4 = b^4 = c^4 &= \frac{64}{9} \\ &&\implies a^4 + b^4 + c^4 &= \frac{64}{3}. \end{align*}$

Thus, the requested inequality holds for any $a,b,c \in \mathbb{R}$ , and as a bonus, we prove equality holds if and only if $a=b=c. \qquad \blacksquare$

Stumbling Robot

Use the pth power mean to prove an inequality

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