Question

Cho a, b, c > 0 và ab + bc + ca = 4abc

Tính Min P= \(\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}\)

Answer 1

Lời giải:

Ta có: \(ab+bc+ac=4abc\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=4\)

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}\right)(1+1+1)\geq \left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)^2\) (1)

\(\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)(1+1+1)\geq \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\) (2)

Từ (1)(2) suy ra \(\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}\geq \frac{1}{27}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^4=\frac{4^4}{27}=\frac{256}{27}\)

Vậy \(P_{\min}=\frac{256}{27}\)

Dấu bằng xảy ra khi \(a=b=c=\frac{3}{4}\)

Answer 2

về mặt toán học dấu cộng (+) khác dấu (.)