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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

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