CMR:với mọi a,b 
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EC
1
7 tháng 8 2017
Áp dụng bđt Cosi cho 2 số dương, ta có:
* \(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}=\frac{a^3}{b^2}+a+\frac{b^3}{c^2}+b+\frac{c^3}{a^2}+c-a-b-c\)\(\ge2\sqrt{\frac{a^3}{b^2}.a}+2\sqrt{\frac{b^3}{c^2}.b}+2\sqrt{\frac{c^3}{a^2}.c}-a-b-c\)\(=2.\frac{a^2}{b}+2.\frac{b^2}{c}+2.\frac{c^2}{a}-a-b-c\)
* \(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-a-b-c=\frac{a^2}{b}+b+\frac{b^2}{c}+c+\frac{c^2}{a}+a-2a-2b-2c\)
\(\ge2\sqrt{\frac{a^2}{b}.b}+2\sqrt{\frac{b^2}{c}.c}+2\sqrt{\frac{c^2}{a}.a}-2a-2b-2c=0\)
\(\Rightarrow\)\(2.\frac{a^2}{b}+2.\frac{b^2}{c}+2.\frac{c^2}{a}-a-b-c\ge\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
\(\Rightarrow\)\(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\ge\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
Nếu đúng cho mình nhé.
NC
13 tháng 8 2020
Bài làm:
Ta có: \(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}\)
\(=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
=> đpcm
13 tháng 8 2020
Ta có: \(VT=\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}+\frac{4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{a+2\sqrt{ab}+b-\left(a-2\sqrt{ab}+b\right)+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4b+4\sqrt{ab}}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4\sqrt{b}\left(\sqrt{b}+\sqrt{a}\right)}{2\left(\sqrt{b}+\sqrt{a}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}=VP\)(đpcm)
DT
1
KH
8 tháng 8 2020
Lần sau đặt đúng môn nhé :D
Ta có: \(ab=1\Leftrightarrow b=\frac{1}{a}\)
Theo đề:
\(\frac{1}{a}+\frac{1}{b}+\frac{2}{a+b}=\frac{1}{a}+\frac{1}{\frac{1}{a}}+\frac{2}{a+\frac{1}{a}}=\frac{1}{a}+a+\frac{2}{a+\frac{1}{a}}\)
Áp dụng BĐT AM-GM:
\(\Rightarrow\frac{1}{a}+a\ge2\sqrt{1}=2\left(1\right)\Leftrightarrow\frac{2}{a+\frac{1}{a}}\ge\frac{2}{2}=1\left(2\right)\)
Cộng theo vế `(1)` và `(2)`
\(\Rightarrow\frac{1}{a}+a+\frac{2}{a+\frac{1}{a}}\ge2+1=3\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{2}{a+b}\ge3\)