Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
doan thi khanh linh câm cái mồm đi.đã ngu lại còn thích k
áp dụng co si ta có:
\(\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\ge\frac{2\sqrt{bc}}{\sqrt{a}}+\frac{2\sqrt{ca}}{\sqrt{b}}+\frac{2\sqrt{ab}}{\sqrt{c}}\)
\(=\left(\frac{\sqrt{bc}}{\sqrt{a}}+\frac{\sqrt{ca}}{\sqrt{b}}\right)+\left(\frac{\sqrt{ca}}{\sqrt{b}}+\frac{\sqrt{ab}}{\sqrt{c}}\right)+\left(\frac{\sqrt{ab}}{\sqrt{c}}+\frac{\sqrt{bc}}{\sqrt{a}}\right)\)
\(\ge2\sqrt{a}+2\sqrt{b}+2\sqrt{c}=\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
\(\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[3]{\sqrt{abc}}=\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)
\(\Rightarrow Q.E.D\)
Sang học 24 tìm ai tên Perfect Blue nhé t làm bên đó rồi đưa link thì lỗi ==" , tìm tên đăng nhập springtime ấy
Ta co:
\(\sqrt{2\left(b+1\right)}\le\frac{b+3}{2}\Rightarrow\frac{a}{\sqrt{2\left(b+1\right)}}\ge\frac{2a}{b+3}\)
Tuong tu:\(\frac{b}{\sqrt{2\left(c+1\right)}}\ge\frac{2b}{c+3};\frac{c}{\sqrt{2\left(a+1\right)}}\ge\frac{2c}{a+3}\)
\(\Rightarrow\frac{1}{\sqrt{2}}\left(\frac{a}{\sqrt{b+1}}+\frac{b}{\sqrt{c+1}}+\frac{c}{\sqrt{a+1}}\right)\ge2\left(\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{a+3}\right)\)
\(\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{a+3}\)
\(=\frac{a^2}{ab+3a}+\frac{b^2}{bc+3b}+\frac{c^2}{ca+3c}\ge\frac{\left(a+b+c\right)^2}{ab+bc+ca+9}\ge\frac{\left(a+b+c\right)^2}{\frac{\left(a+b+c\right)^2}{3}+9}=\frac{9}{\frac{9}{3}+9}=\frac{3}{4}\)
\(\Rightarrow2\left(\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{a+3}\right)\ge\frac{3}{2}\)
Hay \(\frac{a}{\sqrt{b+1}}+\frac{b}{\sqrt{c+1}}+\frac{c}{\sqrt{a+1}}\ge\frac{3\sqrt{2}}{2}\)
Dau '=' xay ra khi \(a=b=c=3\)
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 \(x=y=z=\frac{1}{3}\Leftrightarrow a=b=c=3\)
1,
\(\frac{a}{1+\frac{b}{a}}+\frac{b}{1+\frac{c}{b}}+\frac{c}{1+\frac{a}{c}}=\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\ge\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\frac{2}{2}=1\left(Q.E.D\right)\)
