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Bạn tham khảo ở đây nhé.

C.hóa \(x+y=1\) và dùng C-S:

\(VT^2\le\frac{2x}{\left(y+1\right)^2}+\frac{2y}{\left(x+1\right)^2}\le\frac{8}{9}=VP^2\)

\(BDT\Leftrightarrow\frac{x}{\left(2-x\right)^2}+\frac{y}{\left(2-y\right)^2}\le\frac{4}{9}\left(1\right)\)

Ta có BĐT phụ \(\frac{x}{\left(2-x\right)^2}\le\frac{20}{27}x-\frac{4}{27}\)

\(\Leftrightarrow-\frac{\left(2x-1\right)^2\left(5x-16\right)}{27\left(x-2\right)^2}\le0\) *Đúng*

Tương tự cho 2 BĐT còn lại rồi cộng theo vế:

\(VT_{\left(1\right)}\le\frac{20}{27}\left(x+y\right)-\frac{4}{27}\cdot2=\frac{4}{9}=VP_{\left(1\right)}\)

"=" khi \(x=y=\frac{1}{2}\)

Ta có \(x+y\le1\Leftrightarrow1-x\ge y>0\Leftrightarrow0< x< 1\)

Giả sử \(x^2-\dfrac{3}{4x}-\dfrac{x}{y}\le-\dfrac{9}{4}\)

\(\Leftrightarrow4x^2+9\le\dfrac{3}{x}+\dfrac{4x}{y}\\ \Leftrightarrow\dfrac{4x}{1-x}+\dfrac{3}{x}\ge4x^2+9\\ \Leftrightarrow\dfrac{4x^2+3\left(1-x\right)-x\left(4x^2+9\right)\left(1-x\right)}{x\left(1-x\right)}\ge0\\ \Leftrightarrow\dfrac{4x^4-4x^3+13x^2-12x+3}{x\left(1-x\right)}\ge0\\ \Leftrightarrow\dfrac{\left(x^2+3\right)\left(2x-1\right)^2}{x\left(1-x\right)}\ge0\)

Vì \(x>0;1-x>0\) nên BĐT trên luôn đúng

Vậy ta được đpcm

Dấu \("="\Leftrightarrow x=y=\dfrac{1}{2}\)

\(VT\le\dfrac{x}{2x+2y+2}+\dfrac{y}{2yz+2z+2}+\dfrac{z}{2z+2x+2}\)

Nên ta chỉ cần chứng minh: \(\dfrac{x}{x+y+1}+\dfrac{y}{y+z+1}+\dfrac{z}{z+x+1}\le1\)

\(\Leftrightarrow\dfrac{y+1}{x+y+1}+\dfrac{z+1}{y+z+1}+\dfrac{x+1}{z+x+1}\ge2\)

Thật vậy, ta có:

\(VT=\dfrac{\left(x+1\right)^2}{\left(x+1\right)\left(z+x+1\right)}+\dfrac{\left(y+1\right)^2}{\left(y+1\right)\left(x+y+1\right)}+\dfrac{\left(z+1\right)^2}{\left(z+1\right)\left(y+z+1\right)}\)

\(VT\ge\dfrac{\left(x+y+z+3\right)^2}{\left(x^2+y^2+z^2\right)+3\left(x+y+z\right)+xy+yz+zx+3}\)

\(VT\ge\dfrac{6\left(x+y+z\right)+2\left(xy+yz+zx\right)+12}{3\left(x+y+z\right)+xy+yz+zx+6}=2\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=1\)

Với mọi số thực ta luôn có:

`(x-y)^2>=0`

`<=>x^2-2xy+y^2>=0`

`<=>x^2+y^2>=2xy`

`<=>(x+y)^2>=4xy`

`<=>(x+y)^2>=16`

`<=>x+y>=4(đpcm)`

\(\dfrac{1}{x+3}+\dfrac{1}{y+3}=\dfrac{x+3+y+3}{\left(x+3\right)\left(y+3\right)}\)

\(=\dfrac{x+y+6}{3x+3y+13}\)(vì \(xy=4\))

=> \(\dfrac{x+y+6}{3x+3y+13}\)≤\(\dfrac{2}{5}\)

<=> \(5\left(x+y+6\right)\)≤\(2\left(3x+3y+13\right)\)

<=>\(6x+6y+26-5x-5y-30\)≥\(0\)

<=> \(x+y-4\)≥\(0\)

Áp dụng BĐT AM-GM \(\dfrac{a+b}{2}\)≥\(\sqrt{ab}\)

Ta có \(\dfrac{x+y}{2}\)≥\(\sqrt{xy}\)

<=>\(x+y\) ≥ 2\(\sqrt{xy}\)

=>2\(\sqrt{xy}-4\)≥\(0\)

<=> \(4-4\)≥0

<=>0≥0 ( Luôn đúng )

Vậy \(\dfrac{1}{x+3}+\dfrac{1}{y+3}\)≤\(\dfrac{2}{5}\)

vì trong 3 số x,y,z có ít nhất là 2 số cùng dấu

giả sử \(x,y\le0\)\(\Rightarrow z=-\left(x+y\right)\ge0\)

Mà \(-1\le x,y,z\le1\)nên \(x^2\le\left|x\right|;y^4\le\left|y\right|;z^6\le\left|z\right|\)

\(\Rightarrow x^2+y^4+z^6\le\left|x\right|+\left|y\right|+\left|z\right|=-x-y+z=-\left(x+y\right)+z=2z\le2\)

Dấu " = " xảy ra chẳng hạn x = 0 ; y = -1; z = 1