mk hơi vội nên sai 1 số lỗi nhỏ bn tự sửa nhé
\(A=\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\)
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 \(x=y=z=\frac{1}{3}\)
vì sao từ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\) mà ra được \(\frac{81}{\left(x+y+z\right)^2}\)
Trần Thành Phát Nguyễn:áp dụng Bdt \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}=\frac{9^2}{\left(x+y+z\right)^2}=\frac{81}{\left(x+y+z\right)^2}\) nhé
Theo BĐT AM - GM ta dễ có:
\(\sqrt{x^2+\frac{1}{x^2}}=\sqrt{x^2+\frac{1}{81x^2}+\frac{1}{81x^2}+\frac{1}{81x^2}+.....+\frac{1}{81x^2}}\)
\(\ge\sqrt{82\sqrt[82]{\frac{x^2}{81^{81}\cdot x^{162}}}}\) Một cách tương tự ta lần lượt có được:
\(\Rightarrow LHS\ge\sqrt{82}\left(\sqrt[82]{\frac{x^2}{81^{81}\cdot x^{162}}}+\sqrt[82]{\frac{y^2}{81^{81}\cdot y^{162}}}+\sqrt[82]{\frac{z^2}{81^{81}\cdot z^{162}}}\right)\)
\(\ge3\sqrt{82}\cdot\sqrt[3]{\sqrt[82]{\frac{x^2}{81^{81}\cdot x^{162}}}\sqrt[82]{\frac{y^2}{81^{81}\cdot y^{162}}}\sqrt[82]{\frac{z^2}{81^{81}\cdot z^{162}}}}\)
\(=\sqrt[246]{\frac{1}{81^{243}\cdot\left(xyz\right)^{162}}}\ge\sqrt[246]{\frac{1}{81^{243}\cdot\left(\frac{\left(x+y+z\right)^3}{27}\right)^{162}}}\)
Bí rồi :(