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Cho $a,b,c>0$. Chứng minh rằng $(a+b)^{4}+(b+c)^{4}+(c+a)^{4}\geq \frac{5}{2}(a^{4}+b^{4}+c^{4})$

[DOTOANNANG]

BĐT mạnh hơn: 

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Chuẩn hóa: $$a+ b+ c= 3$$,

$$a, b, c= x+ 1, y+ 1, z+ 1$$,

$$x+ y+ z= 0$$, $$x^{3}+ y^{3}+ z^{3}= 3xyz$$

$$\left ( a+ b \right )^{4}+ \left ( b+ c \right )^{4}+ \left ( c+ a \right )^{4}- \frac{4}{7}\left [ a^{4}+ b^{4}+ c^{4}+ \left ( a+ b+ c \right )^{4} \right ]\geq  0$$

$$\Leftrightarrow  x^{4}+ y^{4}+ z^{4}+ 48\left ( x^{2}+ y^{2}+ z^{2} \right )\geq  72xyz$$

Giả sử: $$xy\geq  0$$, $$z= -x- y$$

$$2\left ( x^{2}+ xy+ y^{2} \right )^{2}+ 96\left ( x^{2}+ xy+ y^{2} \right )+ 72xy\left ( x+ y \right )\geq  0$$

$$\Leftrightarrow 2\left ( x^{2}+ xy+ y^{2} \right )^{2}+ 96\left ( x^{2}+ xy+ y^{2} \right )+ 72xy\left ( x+ y \right )$$

$$\geq  18x^{2}y^{2}+ 72\left ( x+ y \right )^{2}+ 72xy\left ( x+ y \right )= 18\left ( 2x+ 2y+ xy \right )^{2}\geq  0$$