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Áp dụng BĐT Bunhia- cốp -xki ta có
\(M=\left(\sqrt{a}+\sqrt{b}\right)^2\le\left(1^2+1^2\right)\left(a+b\right)\le2\)
Vậy maxM =2 \(\Leftrightarrow a=b=\frac{1}{2}\)
\(S=\sqrt{a+b+c}+\sqrt{b+c+d}+\sqrt{c+d+a}+\sqrt{d+a+b}\)
\(\le\frac{a+b+c}{\sqrt{3}}+\frac{\sqrt{3}}{4}+\frac{b+c+d}{\sqrt{3}}+\frac{\sqrt{3}}{4}+\frac{c+d+a}{\sqrt{3}}+\frac{\sqrt{3}}{4}+\frac{d+a+b}{\sqrt{3}}+\frac{\sqrt{3}}{4}\)
\(=\sqrt{3}+\frac{3}{\sqrt{3}}\left(a+b+c+d\right)=2\sqrt{3}\)
Bài 2:
\(\frac{1}{\sqrt[3]{81}}\cdot P=\frac{1}{\sqrt[3]{9\cdot9\cdot\left(a+2b\right)}}+\frac{1}{\sqrt[3]{9\cdot9\cdot\left(b+2c\right)}}+\frac{1}{\sqrt[3]{9\cdot9\cdot\left(c+2a\right)}}\)
\(\ge\frac{3}{a+2b+9+9}+\frac{3}{b+2c+9+9}+\frac{3}{c+2a+9+9}\ge3\left(\frac{9}{3a+3b+3c+54}\right)=\frac{1}{3}\)
\(\Rightarrow P\ge\sqrt[3]{3}\)
Dấu bằng xẩy ra khi a=b=c=3
Bài 1:
\(ab+bc+ca=5abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=5\)
Theo bđt côsi-shaw ta luôn có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\ge\frac{25}{x+y+z+t+k}\)(x=y=z=t=k>0 ) (*)
\(\Leftrightarrow\left(x+y+z+t+k\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\ge25\)
Áp dụng bđt AM-GM ta có:
\(\hept{\begin{cases}x+y+z+t+k\ge5\sqrt[5]{xyztk}\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\ge5\sqrt[5]{\frac{1}{xyztk}}\end{cases}}\)
\(\Rightarrow\left(x+y+z+t+k\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\ge25\)
\(\Rightarrow\)(*) luôn đúng
Từ (*) \(\Rightarrow\frac{1}{25}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\le\frac{1}{x+y+z+t+k}\)
Ta có: \(P=\frac{1}{2a+2b+c}+\frac{1}{a+2b+2c}+\frac{1}{2a+b+2c}\)
Mà \(\frac{1}{2a+2b+c}=\frac{1}{a+a+b+b+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\frac{1}{a+2b+2c}=\frac{1}{a+b+b+c+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)
\(\frac{1}{2a+b+2c}=\frac{1}{a+a+b+c+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)
\(\Rightarrow P\le\frac{1}{25}\left[5.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right]=1\)
\(\Rightarrow P\le1\left(đpcm\right)\)Dấu"="xảy ra khi a=b=c\(=\frac{3}{5}\)
Với 2 số thực x,y>0, ta có:
\(x^3+y^3-x^2y-xy^2=\left(x+y\right)\left(x-y\right)^2\ge0\). Dấu bằng xảy ra \(\Leftrightarrow x=y\).
Do đó: \(x^3+y^3\ge x^2y+xy^2\Leftrightarrow4x^3+4y^3\ge\left(x+y\right)^3\Leftrightarrow x+y\le\sqrt[3]{4x^3+4y^3}\)Áp dụng bđt vừa cm, ta có: \(S=\sqrt[3]{2a+b}+\sqrt[3]{2b+c}+\sqrt[3]{2c+d}+\sqrt[3]{2d+a}\le\sqrt[3]{8a+12b+4c}+\sqrt[3]{8c+12d+4a}\le\sqrt[3]{48a+48b+48c+48d}=\sqrt[3]{48}\)(vì a+b+c+d=1)
Dấu bằng xảy ra\(\Leftrightarrow a=b=c=d=\dfrac{1}{4}\)(vì a+b+c+d=1)
Bn ơi 3x3 + 3y3 vào cả 2 vế thì 4x3 + 4y3 > 3x3 + 3y3 + x2y + xy2 k phải là (x + y)3