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Đặt \(x=\frac{2a}{b+c};y=\frac{2b}{c+a};z=\frac{2c}{a+b}\) Thì bài toán thành chứng minh
\(3\left(\sqrt{\frac{a+b}{2c}}+\sqrt{\frac{b+c}{2a}}+\sqrt{\frac{c+a}{2b}}\right)^2\ge\frac{8\left(a+b+c\right)^3}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Áp dụng holder ta có:
\(\left(\sqrt{\frac{a+b}{2c}}+\sqrt{\frac{b+c}{2a}}+\sqrt{\frac{c+a}{2b}}\right)^2\left(2c\left(a+b\right)^2+2a\left(b+c\right)^2+2b\left(c+a\right)^2\right)\)
\(\ge\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]^3=8\left(a+b+c\right)^3\)
\(\Rightarrow VT\ge3.\frac{8\left(a+b+c\right)^3}{2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2}\)
Từ đây ta cần chứng minh:
\(3.\frac{8\left(a+b+c\right)^3}{2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2}\ge\frac{8\left(a+b+c\right)^3}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\Leftrightarrow2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2\le3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(\Leftrightarrow a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2\ge0\)( đúng )
Vậy có ĐPCM
\(A=\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\)
Áp dụng Bđt MIncopxki ta có:
\(A\ge\sqrt{\left(x+y+\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}\)
\(\ge\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}\)
\(\ge\sqrt{\left(x+y+z\right)^2+\frac{1}{\left(x+y+z\right)^2}+\frac{80}{\left(x+y+z\right)^2}}\)
\(\ge\sqrt{2+80}=\sqrt{82}\)
Dấu = khi \(x=y=z=\frac{1}{3}\)
\(\sqrt{x^2+y^2+y^2}\ge\sqrt{3\sqrt[3]{x^2y^4}}=\sqrt{3}.\sqrt[3]{xy^2}\)
\(\Rightarrow VT\ge\sqrt{3}\left(\frac{\sqrt[3]{xy^2}}{z}+\frac{\sqrt[3]{yz^2}}{x}+\frac{\sqrt[3]{zx^2}}{y}\right)\)
\(\Rightarrow VT\ge3\sqrt{3}\sqrt[3]{\frac{\sqrt[3]{xy^2.yz^2.zx^2}}{xyz}}=3\sqrt{3}.\sqrt[3]{\frac{\sqrt[3]{x^3y^3z^3}}{xyz}}=3\sqrt{3}\)
Dấu "=" xảy ra khi \(x=y=z\)
Ta có: x=y=z \(\Rightarrow\sqrt{\frac{x^2}{y^2+z^2}}+\sqrt{\frac{y^2}{z^2+x^2}}+\sqrt{\frac{z^2}{x^2+y^2}}=\sqrt{\frac{x^2}{2x^2}}+\sqrt{\frac{x^2}{2x^2}}+\sqrt{\frac{x^2}{2x^2}}\)
\(=\sqrt{\frac{1}{2}}+\sqrt{\frac{1}{2}}+\sqrt{\frac{1}{2}}\ge3\sqrt[3]{\frac{1}{2^3}}=\frac{3}{2}\)(BĐT Cosi)
=> Đpcm.
đề ngu lìn :)) cho x=y=z mới ghê :))