Cho \(a,b>0\) và \(c\ne0\) thỏa \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\). CMR: \(\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\)
Cho a>0, b>0, \(c\ne0\) và \(\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\). CMR:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
Cho a, b, c > 0 thỏa mãn a + b = 2c. CMR \(\frac{1}{\sqrt{a}+\sqrt{c}}+\frac{1}{\sqrt{b}+\sqrt{c}}=\frac{2}{\sqrt{a}+\sqrt{b}}\)
Qui đồng chứng minh tương đương là ra
\(a+b=2c\Rightarrow\left\{{}\begin{matrix}c=\frac{a+b}{2}\\a-c=c-b\end{matrix}\right.\)
\(\frac{1}{\sqrt{a}+\sqrt{c}}+\frac{1}{\sqrt{b}+\sqrt{c}}=\frac{\sqrt{a}-\sqrt{c}}{a-c}+\frac{\sqrt{b}-\sqrt{c}}{b-c}=\frac{\sqrt{a}-\sqrt{c}}{a-c}-\frac{\sqrt{b}-\sqrt{c}}{a-c}\)
\(=\frac{\sqrt{a}-\sqrt{b}}{a-c}=\frac{\sqrt{a}-\sqrt{b}}{a-\frac{a+b}{2}}=\frac{2\left(\sqrt{a}-\sqrt{b}\right)}{a-b}=\frac{2}{\sqrt{a}+\sqrt{b}}\)
Cách khác.
Đặt \(x=\frac{1}{\sqrt{a}+\sqrt{c}};y=\frac{1}{\sqrt{b}+\sqrt{c}};z=\frac{1}{\sqrt{a}+\sqrt{b}}\)(*)
Cần chứng minh \(x+y=2z\)
(*)\(\Leftrightarrow\frac{1}{x}=\sqrt{a}+\sqrt{c};\frac{1}{y}=\sqrt{b}+\sqrt{c};\frac{1}{z}=\sqrt{a}+\sqrt{b}\)
Cộng vế :
\(2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\Leftrightarrow2\cdot\left(\frac{1}{x}+\sqrt{a}\right)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\Leftrightarrow a=\frac{1}{4}\cdot\left(\frac{1}{y}+\frac{1}{z}-\frac{1}{x}\right)^2\)
Tương tự :
\(b=\frac{1}{4}\cdot\left(\frac{1}{x}-\frac{1}{y}+\frac{1}{z}\right)^2\)
\(c=\frac{1}{4}\cdot\left(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}\right)^2\)
Theo giả thiết : \(a+b=2c\)
\(\Leftrightarrow\frac{1}{2}\cdot\left(\frac{1}{x}-\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{4}\cdot\left[\left(\frac{1}{y}+\frac{1}{z}-\frac{1}{x}\right)^2+\left(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}\right)^2\right]\)
\(\Leftrightarrow\frac{4}{xy}-\frac{2}{yz}-\frac{2}{zx}=0\)
\(\Leftrightarrow\frac{2}{xy}=\frac{1}{yz}+\frac{1}{zx}\)
\(\Leftrightarrow\frac{2z}{xyz}=\frac{x+y}{xyz}\)
\(\Leftrightarrow2z=x+y\) ( đpcm )
Cho \(a,b>0\&c\ne0\)
CMR \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Leftrightarrow\sqrt{a+b}=\sqrt{ac}+\sqrt{bc}\)
Cho các số dương a, b, c thỏa mãn a+b+c=0. CMR: \(\sqrt{\frac{a}{1-a}}+\sqrt{\frac{b}{1-b}}+\sqrt{\frac{c}{1-c}}>2\)
Số dương thì sao \(a+b+c=0\) được? Chắc là \(a+b+c=1\) mới đúng
Khi đó:
\(VT=\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}=\frac{2a}{2\sqrt{a\left(b+c\right)}}+\frac{2b}{2\sqrt{b\left(a+c\right)}}+\frac{2c}{2\sqrt{c\left(a+b\right)}}\)
\(VT\ge\frac{2a}{a+b+c}+\frac{2b}{a+b+c}+\frac{2c}{a+b+c}=2\)
Dấu "=" không xảy ra nên:
\(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}>2\)
Cho a,b,c>0 thỏa mãn abc=1
CMR B=\(\frac{a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\le\frac{3\sqrt{2}}{2}\)
Cho abc>0 thỏa mãn abc=1.CMR:\(\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}>=\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)
biến dổi tương đương
cộng trừ VT\(\sqrt{a},\sqrt{b},\sqrt{c}\)
Quy đống lên ta có
\(\left(a+b+c\right)\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)-\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)bạn quy đồng lên rùi lm tiep
Cho \(a>0;b>0;c\ne0\) và \(\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\)
Chứng minh rằng \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
Cho a,b,c>0 thỏa mãn ab+bc+ac=1. CMR \(\frac{a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\le\frac{3}{2}\)
\(VT=\frac{a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\)
\(=\frac{a}{\sqrt{a^2+ab+bc+ca}}+\frac{b}{\sqrt{b^2+ab+bc+ca}}+\frac{c}{\sqrt{c^2+ab+bc+ca}}\)
\(=\frac{a}{\sqrt{\left(a+b\right)\left(c+a\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(b+c\right)\left(c+a\right)}}\)
\(=\sqrt{\frac{a}{a+b}.\frac{a}{c+a}}+\sqrt{\frac{b}{a+b}.\frac{b}{b+c}}+\sqrt{\frac{c}{b+c}.\frac{c}{c+a}}\)
\(\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{c+a}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{b+c}+\frac{c}{c+a}\right)\)
\(=\frac{1}{2}.3=\frac{3}{2}\)
Dấu = xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
cho a,b,c > 0 thỏa mãn a+b+c=abc. Cmr:
\(\frac{\sqrt{a^2+1}}{a}+\frac{\sqrt{b^2+1}}{b}-\sqrt{1+c^2}< 1\)