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Sang học 24 tìm ai tên Perfect Blue nhé t làm bên đó rồi đưa link thì lỗi ==" , tìm tên đăng nhập springtime ấy
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\) thì x, y, z > 0; x + y + z = 1. Quy về: \(\sqrt{\frac{1}{x}+\frac{1}{yz}}+\sqrt{\frac{1}{y}+\frac{1}{zx}}+\sqrt{\frac{1}{z}+\frac{1}{xy}}\ge\sqrt{\frac{1}{xyz}}+\sqrt{\frac{1}{x}}+\sqrt{\frac{1}{y}}+\sqrt{\frac{1}{z}}\)
\(\Leftrightarrow\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge1+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)
\(\Leftrightarrow\frac{x}{\sqrt{x+yz}+\sqrt{yz}}+\frac{y}{\sqrt{y+zx}+\sqrt{zx}}+\frac{z}{\sqrt{z+xy}+\sqrt{xy}}\ge1\) (chuyển vế qua nhóm lại rồi liên hợp)
\(\Leftrightarrow\Sigma_{cyc}\frac{x}{\sqrt{x\left(x+y+z\right)+yz}+\sqrt{yz}}\ge1\Leftrightarrow\Sigma_{cyc}\frac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}+\sqrt{yz}}\ge1\)
BĐT này đúng! Thật vậy:
\(VT\ge\Sigma_{cyc}\frac{x}{\frac{\left(x+y\right)+\left(z+z\right)}{2}+\frac{\left(y+z\right)}{2}}=\Sigma_{cyc}\frac{x}{x+y+z}=\frac{x+y+z}{x+y+z}=1\)
Ta có đpcm. Đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\Leftrightarrow a=b=c=3\)
\(a^2\sqrt{a}+b^2\sqrt{b}+c^2\sqrt{c}+\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\)
\(=\left(a^2\sqrt{a}+\frac{1}{\sqrt{a}}\right)+\left(b^2\sqrt{b}+\frac{1}{\sqrt{b}}\right)+\left(c^2\sqrt{c}+\frac{1}{\sqrt{c}}\right)\)
\(\ge2a+2b+2c\ge6\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=6\)
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1,
\(\frac{a}{1+\frac{b}{a}}+\frac{b}{1+\frac{c}{b}}+\frac{c}{1+\frac{a}{c}}=\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\ge\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\frac{2}{2}=1\left(Q.E.D\right)\)
doan thi khanh linh câm cái mồm đi.đã ngu lại còn thích k
áp dụng co si ta có:
\(\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\ge\frac{2\sqrt{bc}}{\sqrt{a}}+\frac{2\sqrt{ca}}{\sqrt{b}}+\frac{2\sqrt{ab}}{\sqrt{c}}\)
\(=\left(\frac{\sqrt{bc}}{\sqrt{a}}+\frac{\sqrt{ca}}{\sqrt{b}}\right)+\left(\frac{\sqrt{ca}}{\sqrt{b}}+\frac{\sqrt{ab}}{\sqrt{c}}\right)+\left(\frac{\sqrt{ab}}{\sqrt{c}}+\frac{\sqrt{bc}}{\sqrt{a}}\right)\)
\(\ge2\sqrt{a}+2\sqrt{b}+2\sqrt{c}=\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
\(\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[3]{\sqrt{abc}}=\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)
\(\Rightarrow Q.E.D\)