Đáp án:
Giải thích các bước giải:
Đặt vế trái của BĐT cần chứng minh là P
Giả thiết tương đương: $\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}=3$
Ta có: $\dfrac{1}{a^4}+\dfrac{1}{a^4}+\dfrac{1}{a^4}+\dfrac{1}{b^4} \geq 4\sqrt[4]{\dfrac{1}{a^{12}b^4}}=\dfrac{4}{a^3b}$
$⇒\dfrac{3}{a^4}+\dfrac{1}{b^4} \geq \dfrac{4}{a^3b}$
Tương tự: $\dfrac{3}{b^4}+\dfrac{1}{c^4} \geq \dfrac{4}{b^3c}$
$\dfrac{3}{c^4}+\dfrac{1}{a^4} \geq \dfrac{4}{c^3a}$
Cộng vế với vế: $\dfrac{4}{a^4}+\dfrac{4}{b^4}+\dfrac{4}{c^4} \geq \dfrac{4}{a^3b}+\dfrac{4}{b^3c}+\dfrac{4}{c^3a}$
$⇒\dfrac{1}{a^3b}+\dfrac{1}{b^3c}+\dfrac{1}{c^3a} \leq 3$
Ta có:
$\dfrac{1}{a^3b+c^2+c^2+1} \leq \dfrac{1}{16}\left(\dfrac{1}{a^3b}+\dfrac{1}{c^2}+\dfrac{1}{c^2} +\dfrac{1}{1}\right)$
$⇔\dfrac{1}{a^3b+2c^2+1} \leq \dfrac{1}{16}\left(\dfrac{1}{a^3b}+\dfrac{2}{c^2}+1 \right)$
Tương tự:
$\dfrac{1}{b^3c+2a^2+1} \leq \dfrac{1}{16}\left(\dfrac{1}{b^3c}+\dfrac{2}{a^2}+1 \right)$
$\dfrac{1}{c^3a+2b^2+1} \leq \dfrac{1}{16}\left(\dfrac{1}{c^3a}+\dfrac{2}{b^2}+1 \right)$
Cộng vế với vế:
$P \leq \dfrac{1}{16}\left(\dfrac{1}{c^3a}+\dfrac{1}{b^3c}+\dfrac{1}{c^3a}+\dfrac{2}{a^2}+\dfrac{2}{b^2}+\dfrac{2}{c^2}+3 \right)$
$⇒P \leq \dfrac{1}{16}\left(\dfrac{2}{a^2}+\dfrac{2}{b^2}+\dfrac{2}{c^2}+6 \right)$
$⇒P \leq \dfrac{1}{16}\left(2\sqrt{3\left(\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}\right)}+6 \right)$
$⇒P \leq \dfrac{12}{16}=\dfrac{3}{4}$
Dấu "=" xảy ra khi $a=b=c=1$
