Đề được sửa lại là: Cho \(x;y;z>0\) sao cho xyz = 1

cm: \(\dfrac{1}{x^2+y+z}+\dfrac{1}{y^2+x+z}+\dfrac{1}{z^2+x+y}\le\dfrac{3}{x+y+z}\)

Áp dụng BĐT bunhiacopxki ta có:\(\left(x^2+y+z\right)\left(1+y+z\right)\ge\left(x+y+z\right)^2\)

\(\Rightarrow\dfrac{1}{x^2+y+z}\le\dfrac{1+y+z}{\left(x+y+z\right)^2}\) (1)

bn tự chứng minh các BĐT tương tự (1) rồi cộng vế theo vế ta có:

VT= \(\dfrac{1}{x^2+y+z}+\dfrac{1}{y^2+x+z}+\dfrac{1}{z^2+x+y}\le\dfrac{3+2\left(x+y+z\right)}{\left(x+y+z\right)^2}\)

Bài toán cm hoàn tất khi \(\dfrac{3+2\left(x+y+z\right)}{\left(x+y+z\right)^2}\le\dfrac{3}{\left(x+y+z\right)}\)

\(\Leftrightarrow3+2\left(x+y+z\right)\le3\left(x+y+z\right)\Leftrightarrow x+y+z\ge3\)

Áp dụng BĐT cauchy cho x;y;z>0 ta có:

\(x+y+z\ge3\sqrt[3]{xyz}=3.\sqrt[3]{1}=3\)

Ta có đpcm

\(a^2+b^2+1\ge ab+a+b\Leftrightarrow2\left(a^2+b^2+1\right)-2\left(ab+a+b\right)\ge0\)\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\) (luôn đúng)

Đẳng thức xảy ra khi và chỉ khi a = b = 1

Giống bài 3 IMO 2017

Ta có: \(x;y;z\ge0\)\(\left(x+y-z\right)^2+x^2y^2\ge0\Leftrightarrow x^2+y^2+z^2+2xy-2yz-2xz+x^2y^2\ge0\)

\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)\le x^2y^2+4xy+4\)\(\Leftrightarrow\left(x+y+z\right)^2\le\left(xy+2\right)^2\)\(\Leftrightarrow x+y+z\le xy+2\)

Từ đó \(P=\dfrac{x}{2+yz}+\dfrac{y}{2+xz}+\dfrac{z}{2+xy}\)

Ta có: \(x+y+z\le xy+2\Rightarrow\dfrac{z}{x+y+z}\ge\dfrac{z}{xy+2}\)

\(\Rightarrow P\le\sum\dfrac{z}{x+y+z}=1\)\(\Rightarrow MaxB=1\)

Đẳng thức xảy ra chẳng hạn x=y=1; z=0

Mà:\(x\left(2+yz\right)\le\left(\dfrac{x^2+2}{2\sqrt{2}}\right).\left(2+\dfrac{y^2+z^2}{2}\right)\) (cauchy)

\(=\dfrac{\left(x^2+2\right)\left(4+y^2+z^2\right)}{4\sqrt{2}}\le\dfrac{\left(x^2+2+4+y^2+z^2\right)^2}{16\sqrt{2}}\)(cauchy) = \(\dfrac{\left(2+2+4\right)^2}{16\sqrt{2}}=2\sqrt{2}\)\(\Rightarrow\dfrac{x}{2+yz}\ge\dfrac{x^2}{2\sqrt{2}}\)

Từ đó \(P\ge\sum\dfrac{x^2}{2\sqrt{2}}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\) (vì \(x^2+y^2+z^2=2\))

Đẳng thức xảy ra chẳng hạn x=y=0; z=\(\sqrt{2}\)

\(Q=\dfrac{xyz}{z^3\left(x+y\right)}+\dfrac{xyz}{x^3\left(y+z\right)}+\dfrac{xyz}{y^3\left(x+z\right)}\)

\(=\dfrac{1}{z^3\left(x+y\right)}+\dfrac{1}{y^3\left(x+z\right)}+\dfrac{1}{x^3\left(y+z\right)}\) (vì xyz = 1)

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

Áp dụng BĐT cauchy schwarz với x,y,z > 0 ta có:

\(Q\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}{2\left(xy+yz+xz\right)}=\dfrac{\left(xy+yz+xz\right)^2}{2\left(xy+yz+xz\right)}=\dfrac{xy+yz+xz}{2}\)Mặt khác theo BĐT cauchy với x;y;z>0 thì

\(xy+yz+xz\ge3\sqrt[3]{x^2y^2z^2}=3\)

Vậy MinQ = \(\dfrac{3}{2}\Leftrightarrow x=y=z=1\)

\(A=\dfrac{1}{x^2+y^2}+\dfrac{2}{xy}+4xy=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{4xy}+4xy+\dfrac{5}{4xy}\)Áp dụng BĐT \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\left(a,b>0\right)\)(bn tự cm BĐT này) và BĐT cauchy ta có:

\(A\ge\dfrac{4}{x^2+2xy+y^2}+2\sqrt{\dfrac{1}{4xy}.4xy}+\dfrac{5}{\left(x+y\right)^2}\)=

\(=\dfrac{4}{\left(x+y\right)^2}+2+\dfrac{5}{\left(x+y\right)^2}\ge4+2+5=11\)(vì x+y\(\le\)1)

Vậy Min A = 11 \(\Leftrightarrow x=y=\dfrac{1}{2}\)