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

BĐT bên trái rất đơn giản, chỉ cần áp dụng:

\(x^3+x^3+y^3\ge3x^2y\) ; tương tự và cộng lại và được

Ta chứng minh BĐT bên phải:

\(\Leftrightarrow x^4+y^4+z^4+2\ge2\left(x^3+y^3+z^3\right)=\left(x+y+z\right)\left(x^3+y^3+z^3\right)\)

\(\Leftrightarrow2\ge x^3\left(y+z\right)+y^3\left(z+x\right)+z^3\left(x+y\right)\)

\(\Leftrightarrow\dfrac{1}{8}\left(x+y+z\right)^4\ge x^3\left(y+z\right)+y^3\left(z+x\right)+z^3\left(x+y\right)\)

Thật vậy, ta có:

\(\dfrac{1}{8}\left(x+y+z\right)^4=\dfrac{1}{8}\left[x^2+y^2+z^2+2\left(xy+yz+zx\right)\right]^2\)

\(\ge\dfrac{1}{8}.4\left(x^2+y^2+z^2\right).2\left(xy+yz+zx\right)=\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)\)

\(=x^3\left(y+z\right)+y^3\left(z+x\right)+z^3\left(x+y\right)+xyz\left(x+y+z\right)\)

\(\ge x^3\left(y+z\right)+y^3\left(z+x\right)+z^3\left(x+y\right)\) (đpcm)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(0;1;1\right)\) và hoán vị

3: \(P=\dfrac{x}{\left(x+y\right)+\left(x+z\right)}+\dfrac{y}{\left(y+z\right)+\left(y+x\right)}+\dfrac{z}{\left(z+x\right)+\left(z+y\right)}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)+\dfrac{1}{4}\left(\dfrac{y}{y+z}+\dfrac{y}{y+x}\right)+\dfrac{1}{4}\left(\dfrac{z}{z+x}+\dfrac{z}{z+y}\right)=\dfrac{3}{2}\).

Đẳng thức xảy ra khi x = y = x = \(\dfrac{1}{3}\).

Với \(0\le x;y\le1\) ta có:

\(\frac{x}{\sqrt{y+3}}+\frac{y}{\sqrt{x+3}}\ge\frac{x}{\sqrt{1+3}}+\frac{y}{\sqrt{1+3}}=\frac{x+y}{2}\)

Dấu "=" xảy ra <=> x = y = 1

Có: \(0\le x;y\le1\)

=> \(0\le x^2\le x\le1;0\le y^2\le y\le1\)

\(\left(\frac{x}{\sqrt{y+3}}+\frac{y}{\sqrt{x+3}}\right)^2\le2\left(\frac{x^2}{y+3}+\frac{y^2}{x+3}\right)\le2\left(\frac{x}{x+y+2}+\frac{y}{x+y+2}\right)\)

\(=2\left(\frac{x+y+2}{x+y+2}-\frac{2}{x+y+2}\right)\le2\left(1-\frac{2}{1+1+2}\right)=1\)

=> \(\sqrt{\frac{x}{\sqrt{y+3}}+\frac{y}{\sqrt{x+3}}}\le1\)

Dấu "=" xảy ra x<=>  = y =1

Lời giải:

Do \(-2\leq x,y\leq 1\) nên:

\(\left\{\begin{matrix} (x+2)(x-1)\leq 0\\ (y+2)(y-1)\leq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x^2+x-2\leq 0\\ y^2+y-2\leq 0\end{matrix}\right.\)

\(\Rightarrow x^2+x-2+y^2+y-2\leq 0\)

\(\Leftrightarrow x^2+y^2\leq 4-(x+y)=4\)

Ta có đpcm.

\(a,\dfrac{x^2+x+2}{\sqrt{x^2+x+1}}=\dfrac{x^2+x+1+1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}+\dfrac{1}{\sqrt{x^2+x+1}}\left(1\right)\)

Áp dụng BĐT cosi: \(\left(1\right)\ge2\sqrt{\sqrt{x^2+x+1}\cdot\dfrac{1}{\sqrt{x^2+x+1}}}=2\)

Dấu \("="\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)