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a.
\(\dfrac{x}{x+\sqrt{3x+yz}}=\dfrac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}=\dfrac{x}{x+\sqrt{\left(x+y\right)\left(z+x\right)}}\le\dfrac{x}{x+\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}}\)
\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}\le\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Tương tự:
\(\dfrac{y}{y+\sqrt{3y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\) ; \(\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Cộng vế:
\(VT\le\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
b.
\(VP=\dfrac{4\left(a+b+c\right)}{2\sqrt{4a\left(a+3b\right)}+2\sqrt{4b\left(b+3c\right)}+2\sqrt{4c\left(c+3a\right)}}\)
\(VP\ge\dfrac{4\left(a+b+c\right)}{4a+a+3b+4b+b+3c+4c+c+3a}\)
\(VP\ge\dfrac{4\left(a+b+c\right)}{8\left(a+b+c\right)}=\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
1) Ta có ĐK: 0 < a,b,c < 1
\(\sqrt{\frac{a}{1-a}}=\frac{a}{\sqrt{a\left(1-a\right)}}\ge2a\) (BĐT AM-GM cho 2 số a và 1-a)
Tương tự, ta có \(\sqrt{\frac{b}{1-b}}=\frac{b}{\sqrt{b\left(1-b\right)}}\ge2b\) và \(\sqrt{\frac{c}{1-c}}=\frac{c}{\sqrt{c\left(1-c\right)}}\ge2c\)
⇒ \(\sqrt{\frac{a}{1-a}}+\sqrt{\frac{b}{1-b}}+\sqrt{\frac{c}{1-c}}\ge2\left(a+b+c\right)=2\)(do a+b+c=1)
Dấu đẳng thức xảy ra \(\Leftrightarrow\) a = b = c = \(\frac{1}{2}\) (không thoả mãn điều kiện a+b+c=1)
Dấu đẳng thức trên không xảy ra được. Vậy ta có bất đẳng thức\(\sqrt{\frac{a}{1-a}}+\sqrt{\frac{b}{1-b}}+\sqrt{\frac{c}{1-c}}>2\)
1) Áp dụng BĐT bun-hi-a-cốp-xki ta có:
\(\left(a+d\right)\left(b+c\right)\ge\left(\sqrt{ab}+\sqrt{cd}\right)^2\)
\(\Leftrightarrow\sqrt{\left(a+d\right)\left(b+c\right)}\ge\sqrt{ab}+\sqrt{cd}\)( vì a,b,c,d dương )
Dấu " = " xảy ra \(\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)
a) Ta có : \(1+x^2=xy+yz+zx+x^2=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(z+x\right)\)
b) \(\Sigma\left(x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\right)=\Sigma\left(x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right).\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\right)\)
\(=\Sigma\left(x\left(y+z\right)\right)=xy+xz+xy+yz+zx+zy=2\left(xy+yz+zx\right)=2\)
Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)
\(x+y+z\ge\frac{x^2+2xy}{2x+y}+\frac{y^2+2yz}{2y+z}+\frac{z^2+2zx}{2z+x}\)
\(\Leftrightarrow x+y+z\ge\frac{3xy}{2x+y}+\frac{3yz}{2y+z}+\frac{3zx}{2z+x}\)
\(\frac{3xy}{2x+y}\le\frac{3}{9}xy\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{3}\left(x+2y\right)\)
\(\Rightarrow\Sigma_{cyc}\frac{3xy}{2x+y}\le\frac{1}{3}\left[\left(x+2y\right)+\left(y+2z\right)+\left(z+2x\right)\right]=x+y+z\)
Dấu "=" xảy ra khi x=y=z
Ta có \(1+x^2=x^2+xy+yz+xz=\left(x+z\right)\left(x+y\right)\)
Khi đó BĐT <=>
\(\frac{1}{\left(x+y\right)\left(x+z\right)}+\frac{1}{\left(y+z\right)\left(x+z\right)}+\frac{1}{\left(x+y\right)\left(y+z\right)}\ge\frac{2}{3}\left(\frac{x}{\sqrt{\left(x+z\right)\left(x+y\right)}}+...\right)\)
<=> \(\frac{x+y+z}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\frac{x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}}{\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}}\right)^3\)
<=>\(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge\frac{1}{3}\left(x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}\right)^3\)
<=> \(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\right)^3\)(1)
Xét \(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge\frac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\)
<=> \(9\left[xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\right]\ge8\left(xy\left(x+y\right)+xz\left(x+z\right)+yz\left(y+z\right)+3xyz\right)\)
<=> \(xy\left(y+x\right)+yz\left(y+z\right)+xz\left(x+z\right)\ge6xyz\)
<=> \(x\left(y-z\right)^2+z\left(x-y\right)^2+y\left(x-z\right)^2\ge0\)luôn đúng
Khi đó (1) <=>
\(\left(x+y+z\right).\frac{2\sqrt{2}}{3}.\sqrt{x+y+z}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+....\right)^3\)
<=> \(\sqrt{2\left(x+y+z\right)}\ge\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\)
Áp dụng buniacopxki cho vế phải ta có
\(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\le\sqrt{\left(x+y+z\right)\left(3-xy-yz-xz\right)}\)
\(=\sqrt{2\left(x+y+z\right)}\)
=> BĐT được CM
Dấu bằng xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)
Câu b) bạn có làm đc k