Đáp án:
Giải thích các bước giải:
Áp dụng BĐT Co-si, ta có:
`a^6+b^4 \ge 2\sqrt{a^6 b^4}=2a^3 b^2`
`⇒ \frac{2a}{a^6+b^4} \le \frac{2a}{2a^3 b^2}=\frac{1}{a^2 b^2}`
`b^6+c^4 \ge 2\sqrt{b^6 c^4}=2b^3 c^2`
`⇒ \frac{2b}{b^6+c^4} \le \frac{2b}{2b^3 c^2}=\frac{1}{b^2 c^2}`
`a^6+c^4 \ge 2\sqrt{a^6 c^4}=2a^3 c^2`
`⇒ \frac{2c}{c^6+a^4} \le \frac{2c}{2c^3 a^2}=\frac{1}{c^2 a^2}`
Cộng theo vế ta được:
`\frac{2a}{a^6+b^4}+\frac{2b}{b^6+c^4}+\frac{2c}{c^6+a^4} \le \frac{1}{a^2 b^2}+\frac{1}{b^2 c^2}+\frac{1}{c^2 a^2}\ (1)`
Ta có: `\frac{1}{a^4}+\frac{1}{b^4} \ge \frac{2}{\sqrt{a^4 c^4}}=\frac{2}{a^2 b^2}`
Tương tự: `\frac{1}{b^4}+\frac{1}{c^4} \ge \frac{2}{\sqrt{b^4 c^4}}=\frac{2}{b^2 c^2}`
`\frac{1}{c^4}+\frac{1}{a^4} \ge \frac{2}{\sqrt{c^4 a^4}}=\frac{2}{c^2 a^2}`
Cộng theo vế:
`\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{b^4}+\frac{1}{c^4}+\frac{1}{c^4}+\frac{1}{a^4} \ge \frac{2}{a^2 b^2}+\frac{2}{b^2 c^2}+\frac{2}{c^2 a^2}`
`⇔ \frac{2}{a^4}+\frac{2}{b^4}+\frac{2}{c^4} \ge \frac{2}{a^2 b^2}+\frac{2}{b^2 c^2}+\frac{2}{c^2 a^2}`
`⇔ \frac{1}{a^2 b^2}+\frac{1}{b^2 c^2}+\frac{1}{c^2 a^2} \le \frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}\ (2)`
Từ `(1),(2)⇒\frac{2a}{a^6+b^4}+\frac{2b}{b^6+c^4}+\frac{2c}{c^6+a^4} \le \frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}`
`⇒` ĐPCM
Dấu "="xảy ra khi `a=b=c=1`
