Ta có bđt : \(\frac{m^2}{n}+\frac{p^2}{q}\ge\frac{\left(m+p\right)^2}{n+q}\)\(\left(m,n,p,q>0\right)\)(1)

Thật vậy \(\left(1\right)\Leftrightarrow\frac{m^2q+p^2n}{nq}\ge\frac{\left(m+p\right)^2}{n+q}\)

                       \(\Leftrightarrow m^2n\left(n+q\right)+p^2n\left(n+q\right)\ge nq\left(m+p\right)^2\)

                      \(\Leftrightarrow............\)(Phá tung ra + chuyển vế)

                      \(\Leftrightarrow\left(mq-pn\right)^2\ge0\)(Luôn đúng)

Áp dụng (1) ta được

\(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\ge\frac{\left(x+y\right)^2}{a+b}+\frac{z^2}{c}\ge\frac{\left(x+y+z\right)^2}{a+b+c}\)(ĐPCM)

Dấu "=" khi \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)

P/S: nếu hỏi tại sao chỗ bđt phụ lại đặt m,n,p,q khó nhìn thì hãy bảo tại cái đề bài đã có a,b,x,y rồi -.-

Áp dụng BĐT Bunhiacopxki:

\(\left[\left(\frac{x}{\sqrt{a}}\right)^2+\left(\frac{y}{\sqrt{b}}\right)^2+\left(\frac{z}{\sqrt{c}}\right)^2\right]\left[\left(\sqrt{a}\right)^2+\left(\sqrt{b}\right)^2+\left(\sqrt{c}\right)^2\right]\)\(\ge\left(x+y+z\right)^2\)

Hay \(\left(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\right)\left(a+b+c\right)\ge\left(x+y+z\right)^2\)

Chia hai vế của BĐT cho (a + b + c),ta có đpcm: \(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\ge\frac{\left(x+y+z\right)^2}{a+b+c}\)

\(2=4\sqrt{xy}+2\sqrt{xz}\le2x+2y+x+z=3x+2y+z\)

Ta có:

\(VT=\dfrac{3yz}{x}+\dfrac{4zx}{y}+\dfrac{5xy}{z}=2\left(\dfrac{xy}{z}+\dfrac{zx}{y}+\dfrac{yz}{x}\right)+\left(\dfrac{yz}{x}+\dfrac{xy}{z}\right)+2\left(\dfrac{zx}{y}+\dfrac{xy}{z}\right)\)

\(VT\ge2\left(x+y+z\right)+2y+4x\)

\(VT\ge2\left(3x+2y+z\right)\ge4\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)

 A=4x(x+y)(x+z)(x+y+z)+y²z²

A=4x(x+y+z)(x+y)(x+z)+y²z²

A=(4x²+4xy+4xz)(x²+xz+xy+yz) +y²z²

A=4(x²+yx+xz)(x²+yz+xz+yz)+y²z²

đặt x²+yz+z=a

=>A=4a(a+yz)+y²z²

A=4a²+4ayz+y²z²

A=(2a+yz)²

MÀ (2a+yz)²\(\ge\)0

=>A \(\ge\)0 với mọi x,y,z thuộc R

\(x^2+y^2+z^2\ge xy-xz+yz\)

\(\Rightarrow2x^2+2y^2+2z^2\ge2xy-2xz+2yz\)

\(\Rightarrow2x^2+2y^2+2z^2-2xy+2xz-2yz\ge0\)

\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(x^2+2xz+z^2\right)+\left(z^2-2yz+y^2\right)\ge0\)

\(\Rightarrow\left(x-y\right)^2+\left(x+z\right)^2+\left(z-y\right)^2\ge0\)( luôn đúng )

\(\Rightarrow x^2+y^2+z^2\ge xy-xz+yz\)( đúng với mọi x,y,z )

Dấu bằng sảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x+z\right)^2=0\\\left(z-y\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x-y=0\\x+z=0\\z-y=0\end{cases}\Rightarrow\hept{\begin{cases}y=x\\x+z=0\\y=z\end{cases}}}}\)

\(\Rightarrow\hept{\begin{cases}x+z=0\\x=z\end{cases}\Rightarrow x=y=z=0}\)

ta có : \(\left\{{}\begin{matrix}x^2+y^2\ge2xy\\y^2+z^2\ge2yz\\z^2+x^2\ge2zx\end{matrix}\right.\)

cộng quế theo quế ta có : \(2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+zx\right)\)

\(\Leftrightarrow x^2+y^2+z^2-xy-yz-zx\ge0\forall x;y;z\left(đpcm\right)\)

a: \(x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

b: \(x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

c: \(=x^2-2\cdot x\cdot\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\forall x,y\ne0\)

a: Ta có: \(x^2-8x+20\)

\(=x^2-8x+16+4\)

\(=\left(x-4\right)^2+4>0\forall x\)

b: Ta có: \(-x^2+6x-19\)

\(=-\left(x^2-6x+19\right)\)

\(=-\left(x^2-6x+9+10\right)\)

\(=-\left(x-3\right)^2-10< 0\forall x\)