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Ta có : \(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2}{b+c}.\frac{b+c}{4}}=a\)
Tương tự : \(\frac{b^2}{a+c}+\frac{a+c}{4}\ge b\) ; \(\frac{c^2}{a+b}+\frac{a+b}{4}\ge c\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\left(a+b+c\right)-\frac{2\left(a+b+c\right)}{4}=\frac{a+b+c}{2}=\frac{3}{2}\)
Vậy Min = 3/2 \(\Leftrightarrow a=b=c=1\)
a:
ĐKXĐ: x<>2
|2x-3|=1
=>\(\left[{}\begin{matrix}2x-3=1\\2x-3=-1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Thay x=1 vào A, ta được:
\(A=\dfrac{1+1^2}{2-1}=\dfrac{2}{1}=2\)
b: ĐKXĐ: \(x\notin\left\{-1;2\right\}\)
\(B=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{x^2-x-2}\)
\(=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{2x\left(x-2\right)+3\left(x+1\right)-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{2x^2-4x+3x+3-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{-x+2}{\left(x+1\right)\left(x-2\right)}=-\dfrac{1}{x+1}\)
c: \(P=A\cdot B=\dfrac{-1}{x+1}\cdot\dfrac{x\left(x+1\right)}{2-x}=\dfrac{x}{x-2}\)
\(=\dfrac{x-2+2}{x-2}=1+\dfrac{2}{x-2}\)
Để P lớn nhất thì \(\dfrac{2}{x-2}\) max
=>x-2=1
=>x=3(nhận)
Áp dụng bất đẳng thức cho hai số dương
\(\dfrac{1}{\left(a+b\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Xét \(c+1=c+a+b+c\)
\(\dfrac{ab}{c+1}\le\dfrac{ab}{4\left[\dfrac{1}{a+c}+\dfrac{1}{b+c}\right]}\)
Tương tự:
\(\dfrac{bc}{a+1}\le\dfrac{bc}{4\left[\dfrac{1}{a+c}+\dfrac{1}{b+a}\right]}\)
\(\dfrac{ca}{b+1}\le\dfrac{ac}{4\left[\dfrac{1}{a+b}+\dfrac{1}{c+b}\right]}\)
Cộng lại :
\(\dfrac{ab}{c+1}+\dfrac{bc}{a+1}+\dfrac{ca}{b+1}\le\dfrac{1}{4}\left[\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{bc}{a+c}+\dfrac{bc}{a+b}+\dfrac{ac}{a+b}+\dfrac{ac}{b+c}\right]\)
Rút gọn mẫu số
\(\Rightarrow\dfrac{ab}{c+1}+\dfrac{bc}{a+1}+\dfrac{ca}{b+1}\le\dfrac{1}{4}\left(a+b+c\right)=\dfrac{1}{4}\)
