Chứng minh bằng phép biến đổi tương đương:

1.

\(\Leftrightarrow4+x+y\ge4\sqrt{x+y}\)

\(\Leftrightarrow x+y-4\sqrt{x+y}+4\ge0\)

\(\Leftrightarrow\left(\sqrt{x+y}-2\right)^2\ge0\) (luôn đúng)

Vậy BĐT đã cho đúng

2.

\(\Leftrightarrow\dfrac{y+z}{xyz}\ge\dfrac{4}{x^2+yz}\)

\(\Leftrightarrow\left(y+z\right)\left(x^2+yz\right)\ge4xyz\)

\(\Leftrightarrow x^2y+x^2z+y^2z+z^2y-4xyz\ge0\)

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

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

áp dụng bđt dang Engel

P=1/[x(x+y) ]+1/[y(x+y) ]

=1/(x+y). (1/x+1/y)

=1/(x+y). [(x+y) /xy]=1/(xy)

x+y≤1,x, y>0=>x.y≤1/4

p≥1/(1/4)=4

đẳng thức khi x=y=1/2

 Trước hết, ta đi chứng minh một bổ đề sau: Nếu \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\). Thật vậy, ta phân tích 

 \(P=a^3+b^3+c^3-3abc\)

\(P=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(P=\left(a+b+c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(P=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\).

Hiển nhiên nếu \(a+b+c=0\) thì \(P=0\) hay \(a^3+b^3+c^3=3abc\), bổ đề được chứng minh.

Do \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\) nên áp dụng bổ đề, ta được \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\).

Vì vậy \(\dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{xy}{z^2}=\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}\) \(=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)\) \(=xyz.\dfrac{3}{xyz}=3\). Ta có đpcm

\(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}\ge\dfrac{2}{1+xy}\)

⇔ \(\left(\dfrac{1}{1+x^2}-\dfrac{1}{1+xy}\right)+\left(\dfrac{1}{1+y^2}-\dfrac{1}{1+xy}\right)\ge0\)

⇔ \(\left(\dfrac{1+xy-\left(1+x^2\right)}{\left(1+x^2\right)\left(1+xy\right)}\right)+\left(\dfrac{1+xy-\left(1+y^2\right)}{\left(1+y^2\right)\left(1+xy\right)}\right)\ge0\)

⇔ \(\left(\dfrac{1+xy-1-x^2}{\left(1+x^2\right)\left(1+xy\right)}\right)+\left(\dfrac{1+xy-1-y^2}{\left(1+y^2\right)\left(1+xy\right)}\right)\ge0\)

⇔ \(\dfrac{-x\left(x-y\right)}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{-y\left(y-x\right)}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)

⇔ \(\dfrac{-x\left(x-y\right)\left(1+y^2\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}+\dfrac{y\left(x-y\right)\left(1+x^2\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)

=> -x(x-y)(1+y²)+y(x-y)(1+x²) ≥ 0

⇔ (x-y)[-x(1+y²)+y(1+x²)]≥0

⇔ (x-y)(-x-xy²+y+x²y) ≥0

⇔ (x-y)[-(x-y)+(x²y-y²x)] ≥ 0

⇔ (x-y)[-(x-y)+xy(x-y) ]≥ 0

⇔ (x-y)(x-y)(xy-1)≥ 0

⇔ (x-y)² (xy-1) ≥0 (luôn đúng ∀ xy ≥ 1)

=> đpcm

Áp dụng BĐT AM-GM:

\(VT=\sum\dfrac{\sqrt{\left(x+y\right)^2-xy}}{4yz+1}\ge\sum\dfrac{\sqrt{\left(x+y\right)^2-\dfrac{1}{4}\left(x+y\right)^2}}{\left(y+z\right)^2+1}=\sum\dfrac{\dfrac{\sqrt{3}}{2}\left(x+y\right)}{\left(y+z\right)^2+1}\)

Set \(\left\{{}\begin{matrix}x+y=a\\y+z=b\\z+x=c\end{matrix}\right.\)thì giả thiết trở thành \(a+b+c=3\) và cần chứng minh \(\dfrac{\sqrt{3}}{2}.\sum\dfrac{a}{b^2+1}\ge\dfrac{3\sqrt{3}}{4}\)

\(\Leftrightarrow\sum\dfrac{a}{b^2+1}\ge\dfrac{3}{2}\)( đến đây quen thuộc rồi)

Ta có:\(\sum\dfrac{a}{b^2+1}=\sum a-\sum\dfrac{ab^2}{b^2+1}\ge3-\sum\dfrac{ab^2}{2b}\)(AM-GM)

\(VT\ge3-\sum\dfrac{ab}{2}\ge3-\dfrac{\dfrac{1}{3}\left(a+b+c\right)^2}{2}=\dfrac{3}{2}\)( AM-GM)

Vậy ta có đpcm.Dấu = xảy ra khi a=b=c=1 hay \(x=y=z=\dfrac{1}{2}\)

\(VT=\dfrac{3}{xy+yz+xz}+\dfrac{2}{x^2+y^2+z^2}\)

\(=\dfrac{8}{4\left(xy+yz+xz\right)}+\dfrac{4}{4\left(xy+yz+xz\right)}+\dfrac{4}{2\left(x^2+y^2+z^2\right)}\)

\(\ge\dfrac{8}{4\cdot\dfrac{\left(x+y+z\right)^2}{3}}+\dfrac{\left(2+2\right)^2}{2\left(x+y+z\right)^2}\)

\(=\dfrac{8}{4\cdot\dfrac{1^2}{3}}+\dfrac{\left(2+2\right)^2}{2\cdot1^2}=14\)

\("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)

cái chỗ math processing error kia là \(3\left(\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}+\dfrac{1}{z^2+1}\right)+\left(1+x^2\right)\left(1+y^2\right)\left(1+z^2\right)\ge\dfrac{985}{108}\)

\(H=\dfrac{1}{1+xy}+\dfrac{1}{1+yz}+\dfrac{1}{1+xz}\ge\dfrac{\left(1+1+1\right)^2}{3+xy+yz+xz}=\dfrac{9}{3+xy+yz+xz}\)

Mặt khác,theo AM-GM: \(xy+yz+xz\le x^2+y^2+z^2=3\)

\(\Rightarrow\dfrac{9}{3+xy+yz+xz}\ge\dfrac{9}{3+3}=\dfrac{9}{6}=\dfrac{3}{2}\)

Dấu "=" khi: \(x=y=z=1\)