$ My $ $ solution $ $ a^{12}+b^{12}+c^{12}+4(a+b+c)^2\geq 4(a^2+b^2+c^2)+27 \implies a^{12}+b^{12}+c^{12}-4(a^2+b^2+c^2)+9\geq 0 $ $ Lagrange $ $ method $ $ f=a^{12}+b^{12}+c^{12}-4(a^2+b^2+c^2)+9 $ $ g=a+b+c-3=0 $ $ L=f-\lambda g $ $\frac{d(L)}{d(a)}=0\implies \lambda =12a^{11}-8a $ $\frac{d(L)}{d(b)}=0\implies \lambda =12b^{11}-8b $ $\frac{d(L)}{d(c)}=0\implies \lambda =12c^{11}-8c $ $ Then $ $ 3a^{11}-2a=3b^{11}-2b $ $ K(a)=3a^{11}-2a $ $ K''(a)\geq 0 \implies K(a)-increasing $ $ Then $ $ a=b=c=1 $ $ Then $ $ f(1,1,1)=1^{12}+1^{12}+1^{12}-4(1^2+1^2+1^2)+9=0 $

By AM-GM we have $a^{12}+1+1+1+1+1\geq 6a^2$ so \begin{align*} LHS &\geq 6(a^2+b^2+c^2)-15+8(ab+bc+ca) \\ &= 2(a^2+b^2+c^2)+21 \end{align*}so we just have to prove that $a^2+b^2+c^2\geq 3$, which is easy.

Let $a, b, c,d $ are real positive numbers such that $a+b+c+d=4.$ Prove that $$a^{12}+b^{12}+c^{12}+d^{12}+8(ab+ac+ad+bc+bd+cd)\geq 52 $$Let $a, b, c$ are real positive numbers such that $a+b+c=3.$ Prove that $$a^4+b^4+c^4+8(ab+bc+ca)\geq 27$$Let $a, b, c,d $ are real positive numbers such that $a+b+c+d=4.$ Prove that $$a^4+b^4+c^4+d^4+8(ab+ac+ad+bc+bd+cd)\geq 52 $$ VicKmath7 wrote: $a, b, c$ are real positive numbers for which $a+b+c=3$. Prove that $$a^{12}+b^{12}+c^{12}+8(ab+bc+ca)\geq 27$$

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