Dự đoán dấu = xảy ra khi x=y=\(\dfrac{z}{2}\)

ta có: \(VT=3+\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+\dfrac{y^2}{z^2}+\dfrac{z^2}{y^2}+\dfrac{x^2}{z^2}+\dfrac{z^2}{x^2}\)

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

Áp dụng BĐT AM-GM: \(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\ge2\)

Áp dụng BĐT bunyakovsky:\(\dfrac{y^2}{z^2}+\dfrac{x^2}{z^2}\ge\dfrac{1}{2}\left(\dfrac{y}{z}+\dfrac{x}{z}\right)^2=\dfrac{1}{2}.\dfrac{\left(x+y\right)^2}{z^2}\)

\(\dfrac{z^2}{x^2}+\dfrac{z^2}{y^2}\ge\dfrac{1}{2}\left(\dfrac{z}{x}+\dfrac{z}{y}\right)^2\ge\dfrac{1}{2}\left(\dfrac{4z}{x+y}\right)^2=\dfrac{8z^2}{\left(x+y\right)^2}\)(AM-GM)

do đó \(VT\ge5+\dfrac{1}{2}\dfrac{\left(x+y\right)^2}{z^2}+\dfrac{8z^2}{\left(x+y\right)^2}\)

Đặt \(\dfrac{z}{x+y}=a\)(a>0)thì \(a\ge1\)do \(z\ge x+y\)

\(VT\ge8a^2+\dfrac{1}{2a^2}+5=\dfrac{a^2}{2}+\dfrac{1}{2a^2}+\dfrac{15}{2}a^2+5\ge\dfrac{a^2}{2}+\dfrac{1}{2a^2}+\dfrac{25}{2}\)

Áp dụng BĐT AM-GM: \(\dfrac{a^2}{2}+\dfrac{1}{2a^2}\ge2\sqrt{\dfrac{a^2}{4a^2}}=1\)

do đó \(VT\ge1+\dfrac{25}{2}=\dfrac{27}{2}\)(đpcm)

Dấu = xảy ra khi a=1 hay \(x=y=\dfrac{z}{2}\)

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

Áp dụng Bất đẳng thức AM-GM cho 4 số dương :

\(\Rightarrow2x+xy+z+yzt\ge4\sqrt[4]{2x^2y^2z^2t}\)

\(\Rightarrow1\ge4\sqrt[4]{2x^2y^2z^2t}\Rightarrow1\ge512.x^2y^2z^2t\Rightarrow x^2y^2z^2t\le\dfrac{1}{512}\)

=> MaxI=\(\dfrac{1}{152}\) khi \(\left\{{}\begin{matrix}x=\dfrac{1}{8}\\y=2\\z=\dfrac{1}{4}\\t=\dfrac{1}{2}\end{matrix}\right.\)

Hà Nam Phan Đình cho tớ hỏi BĐT AM-GM là BĐT gì vậy? và lớp mấy được hok vậy ạ?

Xét : A = x^2017+x^2017+1+1+.....+1 ( 2015 số 1 )

Áp dụng bđt cosi thì : 

A >= \(2017\sqrt[2017]{x^{2017}.x^{2017}}\) = 2017.x^2

=> x^2 < = 2x^2017+2015/2017

Tương tự : y^2 < = 2y^2017+2015/2017 ; z^2 < = 2z^2017+2015/2017

=> x^2+y^2+z^2 < = 2(x^2017+y^2017+z^2017)+6045/2017 = 2.3+6045/2017 = 3

Dấu "=" xảy ra <=> x=y=z=1

Vậy GTLN của x^2+y^2+z^2 = 3 <=> x=y=z=1

Tk mk nha