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\(a^2+ab+b^2=\dfrac{1}{2}\left(a+b\right)^2+\dfrac{1}{2}\left(a^2+b^2\right)\ge\dfrac{1}{2}\left(a+b\right)^2+\dfrac{1}{4}\left(a+b\right)^2=\dfrac{3}{4}\left(a+b\right)^2\)
Tương tự, ta có:
\(M\ge\dfrac{\sqrt{3}}{2}\left(a+b\right)+\dfrac{\sqrt{3}}{2}\left(b+c\right)+\dfrac{\sqrt{3}}{2}\left(c+a\right)=\sqrt{3}\left(a+b+c\right)=3\sqrt{3}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(\dfrac{\sqrt{ab+2c^2}}{\sqrt{1+ab-c^2}}=\dfrac{\sqrt{ab+2c^2}}{\sqrt{a^2+b^2+ab}}=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+2c^2\right)}}\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)
\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+a^2+b^2+2c^2}=\dfrac{ab+2c^2}{a^2+b^2+c^2}=ab+2c^2\)
Tương tự và cộng lại:
\(VT\ge ab+bc+ca+2\left(a^2+b^2+c^2\right)=2+ab+bc+ca\)
Áp dụng BĐT Cauchy swarchz ta có:
A=\(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{a+c}\ge\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2} \)
Mà \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=1 \)
=>\(A\ge\frac{1}{2} \)
Dấu "=" xảy ra <=>a=b=c=\(\frac{1}{3} \)
Ta có : \(\sqrt{2a^2+ab+b^2}=\sqrt{\frac{5}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2}\)
Vì \(\frac{3}{4}\left(a-b\right)^2\ge0\forall a;b\Rightarrow\sqrt{2a^2+ab+b^2}\ge\sqrt{\frac{5}{4}}\left(a+b\right)\)( 1 )
Tương tự , ta có : \(\sqrt{2b^2+bc+c^2}\ge\sqrt{\frac{5}{4}}\left(b+c\right);\sqrt{2c^2+ac+a^2}\ge\sqrt{\frac{5}{4}}\left(a+c\right)\left(2\right)\)
Từ ( 1 ) ; ( 2 ) \(\Rightarrow P\ge\sqrt{\frac{5}{4}}.2\left(a+b+c\right)=\sqrt{5}\left(a+b+c\right)\)
Áp dụng BĐT phụ \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\) , ta có :
\(P\ge\sqrt{5}.\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{3}=\frac{\sqrt{5}}{3}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=\frac{1}{9}\)
Ta có :
\(\frac{a^2}{a+b}=\frac{a\left(a+b\right)-ab}{a+b}=a-\frac{ab}{a+b}\text{≥}a-\frac{ab}{2\sqrt{ab}}=a-\frac{\sqrt{ab}}{2}\)(1)
Tương tự : \(\hept{\begin{cases}\frac{b^2}{b+c}\text{≥}b-\frac{\sqrt{bc}}{2}\left(2\right)\\\frac{c^2}{c+a}\text{≥}c-\frac{\sqrt{ac}}{2}\left(3\right)\end{cases}}\)
Cộng vế với vế của (1);(2)(;(3) lại ta được :
\(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{a+c}\text{≥}a+b+c-\frac{\sqrt{ab}}{2}-\frac{\sqrt{bc}}{2}-\frac{\sqrt{ac}}{2}\)
\(\Leftrightarrow A\text{≥}\left(a+b+c-\sqrt{ab}-\sqrt{bc}-\sqrt{ab}\right)+\left(\frac{\sqrt{ab}}{2}+\frac{\sqrt{bc}}{2}+\frac{\sqrt{ac}}{2}\right)\)
Lại lại có : \(a+b+c\text{≥}\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\) (tự chứng minh)
\(\Rightarrow a+b+c-\sqrt{ab}-\sqrt{bc}-\sqrt{ab}\text{≥}0\)
Nên \(A\text{≥}\frac{1}{2}\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)=\frac{1}{2}\)có GTNN là 1/2
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
\(\sqrt{2}M=\sqrt{\left(a-b\right)^2+\left(a^2+b^2\right)}+\sqrt{\left(b-c\right)^2+\left(b^2+c^2\right)}+\sqrt{\left(c-a\right)^2+\left(c^2+a^2\right)}\ge\sqrt{2ab}+\sqrt{2bc}+\sqrt{2ca}\)\(\Leftrightarrow M\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)
Dấu bằng xảy ra khi và chỉ khi a = b, b = c, c = a \(\Leftrightarrow\)a = b = c = \(\frac{1}{3}\)(vì a + b + c = 1).
Suy ra : \(M\ge\sqrt{\frac{1}{3}.\frac{1}{3}}+\sqrt{\frac{1}{3}.\frac{1}{3}}+\sqrt{\frac{1}{3}.\frac{1}{3}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\)
Vậy GTNN của M là 1 khi a = b = c = \(\frac{1}{3}\)