Question

cho a,b,c >0 CMR:

\(\frac{a^4+b^4+c^4}{abc}\ge a+b+c\)

Answer 1

Cách khác:

Xét hiệu:\(a^4+b^4+c^4-abc\left(a+b+c\right)\)

\(=\frac{1}{4}\left[\left(a^2+c^2-2b^2\right)^2+\left(ab+bc-2ca\right)^2\right]+\frac{3}{4}\left(a-c\right)^2\left[\left(a+c\right)^2+b^2\right]\ge0\)

Dấu "=" xảy ra khi \(a=b=c\)

P/s: Bài đơn giản, làm 3 dòng:DDD (vắn tắt tuyệt đối)

Answer 2

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\((a^4+b^4+c^4)(1+1+1)\geq (a^2+b^2+c^2)^2\)

\((a^2+b^2+c^2)(1+1+1)\geq (a+b+c)^2\)

\(\Rightarrow 3(a^4+b^4+c^4)\geq (a^2+b^2+c^2).\frac{(a+b+c)^2}{3}\)

\(\Leftrightarrow a^4+b^4+c^4\geq \frac{(a^2+b^2+c^2)(a+b+c)}{9}.(a+b+c)(1)\)

Áp dụng BĐT AM-GM:

\((a^2+b^2+c^2)(a+b+c)\geq 3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}=9abc(2)\)

Từ $(1);(2)\Rightarrow a^4+b^4+c^4\geq abc(a+b+c)$

hay $\frac{a^4+b^4+c^4}{abc}\geq a+b+c$ (đpcm)

Dấu "=" xảy ra khi $a=b=c$

Answer 3

Cách khác:

Xét hiệu:

\(a^4+b^4+c^4-abc(a+b+c)=\frac{2a^4+2b^4+2c^4-2abc(a+b+c)}{2}\)

\(=\frac{(a^4-2a^2b^2+b^4)+(b^4-2b^2c^2+c^4)+(c^4-2c^2a^2+a^4)+2a^2b^2+2b^2c^2+2c^2a^2-2abc(a+b+c)}{2}\)

\(=\frac{(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2+(a^2b^2-2ab^2c+b^2c^2)+(b^2c^2-2abc^2+c^2a^2)+(a^2b^2-2a^2bc+c^2a^2)}{2}\)

\(=\frac{(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2+(ab-bc)^2+(bc-ac)^2+(ab-ac)^2}{2}\geq 0, \forall a,b,c>0\)

Do đó $a^4+b^4+c^4\geq abc(a+b+c)$

$\Rightarrow \frac{a^4+b^4+c^4}{abc}\geq a+b+c$ (đpcm)