\(1=\left(x+y+z\right)^2\ge4x\left(y+z\right)\Rightarrow x\le\frac{1}{4\left(y+z\right)}\)

Do đó \(A\ge y+z-16yz.\frac{1}{4\left(y+z\right)}+2017\)

\(=y+z-\frac{4yz}{y+z}+2017\ge y+z-\frac{\left(y+z\right)^2}{y+z}+2017=2017\)

Đẳng thức xảy ra khi \(\left(x;y;z\right)=\left(\frac{1}{2};\frac{1}{4};\frac{1}{4}\right)\)

Is that true?

Bí quá thì làm cách sau đây cũng được:v

Đặt A= f(x;y;z) và \(t=\frac{y+z}{2}>0\). Xét hiệu:

\(f\left(x;y;z\right)-f\left(x;t;t\right)=16x\left(t^2-yz\right)\ge0\)

Do đó \(f\left(x;y;z\right)\ge f\left(x;t;t\right)=f\left(1-2t;t;t\right)\) (do cách chọn t)

Ta sẽ tìm min của \(f\left(1-2t;t;t\right)=2t-16\left(1-2t\right)t^2+2017\)

\(=2t\left(4t-1\right)^2+2017\ge2017\)

Đẳng thức xảy ra khi \(y=z=t=\frac{1}{4}\Rightarrow x=\frac{1}{2}\)

\(x\left(x-z\right)+y\left(y-z\right)=0\)\(\Leftrightarrow\)\(x^2+y^2=z\left(x+y\right)\)

\(\frac{x^3}{z^2+x^2}=x-\frac{z^2x}{z^2+x^2}\ge x-\frac{z^2x}{2zx}=x-\frac{z}{2}\)

\(\frac{y^3}{y^2+z^2}=y-\frac{yz^2}{y^2+z^2}\ge y-\frac{yz^2}{2yz}=y-\frac{z}{2}\)

\(\frac{x^2+y^2+4}{x+y}=\frac{z\left(x+y\right)+4}{x+y}=z-x-y+\frac{4}{x+y}+x+y\ge z-x-y+4\)

Cộng lại ra minP=4, dấu "=" xảy ra khi \(x=y=z=1\)

\(x,y,z>0\)

Áp dụng BĐT Caushy cho 3 số ta có:

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

\(P=\dfrac{x^3-1}{x^2+y+z}+\dfrac{y^3-1}{x+y^2+z}+\dfrac{z^3-1}{x+y+z^2}\)

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

Áp dụng BĐT Caushy-Schwarz ta có:

\(P\ge\dfrac{\left(x^3+y^3+z^3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}\)

\(\ge\dfrac{\left(3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}=0\)

\(P=0\Leftrightarrow x=y=z=1\)

Vậy \(P_{min}=0\)

Ta có : \(A^2=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+\frac{2x\sqrt{y}}{\sqrt{z}}+\frac{2y\sqrt{z}}{\sqrt{x}}+\frac{2z\sqrt{x}}{\sqrt{y}}\)

Áp dụng BĐT Cô-si cho 4 số dương,ta có ;

\(\frac{x^2}{y}+\frac{x\sqrt{y}}{\sqrt{z}}+\frac{x\sqrt{y}}{\sqrt{z}}+z\ge4\sqrt[4]{\frac{x^2.x^2.y.z}{yz}}=4x\)

Tương tự : ....

\(\Rightarrow A^2\ge4\left(x+y+z\right)-\left(x+y+z\right)=3\left(x+y+z\right)\ge36\)

\(\Rightarrow A\ge6\)

Dấu "=" xảy ra khi x = y = z = 4

Đặt \(\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)\rightarrow\left(a;b;c\right)\)

Khi đó \(a^2+b^2+c^2\ge12\) ta cần tìm GTNN của  \(A=\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)

\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge2\sqrt{\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\left(a+b+c\right)}\)

Ta có:\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2b+b^2c+c^2a}\)

Mà \(\left(a^2+b^2+c^2\right)\left(a+b+c\right)\ge3\left(a^2+b^2+c^2\right)\) ( cơ bản )

\(\Rightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge2\sqrt{3\left(a^2+b^2+c^2\right)}=12\)

\(\Rightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge12-\left(a+b+c\right)\)

Chứng minh được \(a+b+c\le6\) là OKE nhưng có vẻ không ổn lắm :))

<=> A = (x+y) + ( 5/x + 5/y) +( 25/x + x)

Xét:

+) x+y >/ 10

+) 5/x + 5/y = 5(1/x+1/y) >/ 5.4/x+y = 2 <=> x=y

+) 25/x + x >/ 2. căn 25/x.x =10

=> A >/ 10+2+10 = 22 <=> (x;y)= (5;5).

\(A=\left(\dfrac{6x}{5}+\dfrac{30}{x}\right)+\left(\dfrac{y}{5}+\dfrac{5}{y}\right)+\dfrac{4}{5}\left(x+y\right)\)

\(A\ge2\sqrt{\dfrac{180x}{5x}}+2\sqrt{\dfrac{5y}{5y}}+\dfrac{4}{5}.10=22\)

\(A_{min}=22\) khi \(x=y=5\)

Tham khảo link này nha

https://olm.vn/hoi-dap/detail/243232541423.htm

Ta có:

\(4A=\frac{\left(x+y+z+t\right)^2\left(x+y+z\right)\left(x+y\right)}{xyzt}\)

\(\ge\frac{4\left(x+y+z\right)t\left(x+y+z\right)\left(x+y\right)}{xyzt}\)

\(=\frac{4\left(x+y+z\right)^2\left(x+y\right)}{xyz}\ge\frac{16\left(x+y\right)z\left(x+y\right)}{xyz}\)

\(=\frac{16\left(x+y\right)^2}{xy}\ge\frac{64xy}{xy}=64\)

\(\Rightarrow A\ge16\)

Đấu = xảy ra khi \(t=2z=4x=4y=1\)

x;y;z;t >0 áp dụng bất đẳng thức Cô-si cho 2 số dương ta có :

=\(x+y\ge2\sqrt{xy}\)

=\(\left(x+y\right)+z\ge2\sqrt{\left(x+y\right)z}\)

=\(\left(x+y+z\right)+t\ge2\sqrt{\left(x+y+z\right)t}\)

nhân các vế tương ứng ta có:

\(\left(x+y\right)\left(x+y+z\right)\left(x+y+z+t\right)\ge8\sqrt{xyzt\left(x+y\right)\left(x+y+z\right)}\)

mà x+y+z+t=2

\(\left(x+y\right)\left(x+y+z\right)2\ge8\sqrt{xyzt\left(x+y\right)\left(x+y+z\right)}\)

=\(\sqrt{\left(x+y\right)\left(x+y+z\right)}\ge4\sqrt{xyzt}\)

=\(\left(x+y\right)\left(x+y+z\right)\ge16xyzt\)

\(\Rightarrow B=\frac{\left(x+y\right)\left(x+y+z\right)}{xyzt}\ge\frac{16xyzt}{xyzt}=16\)

vậy minB=16 khi\(\hept{\begin{cases}x=y\\x+y=z\\x+y+z=t\end{cases}};x+y+z+t=2\Rightarrow x=y=0.25;z=0.5;t=1\)