\(A=\dfrac{1}{z}\left(\dfrac{x+y}{xy}\right)=\dfrac{1}{z}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge\dfrac{4}{z\left(x+y\right)}\ge\dfrac{16}{\left(x+y+z\right)^2}=16\)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(\dfrac{1}{4};\dfrac{1}{4};\dfrac{1}{2}\right)\)

Áp dụng BĐT Cô-si cho 2 số thực dương \(\dfrac{xy}{z}\) và \(\dfrac{yz}{x}\) có:

\(\dfrac{xy}{z}+\dfrac{yz}{x}\) \(\ge\) 2\(\sqrt{\dfrac{xy}{z}\cdot\dfrac{yz}{x}}\) = 2\(\sqrt{y^2}\) = 2y (1)

Tương tự: \(\dfrac{yz}{x}+\dfrac{zx}{y}\ge2z\) (2)

\(\dfrac{xy}{z}+\dfrac{zx}{y}\ge2x\) (3)

Từ (1); (2); (3)

\(\Rightarrow\) \(\dfrac{2xy}{z}+\dfrac{2yz}{x}+\dfrac{2zx}{y}\ge2x+2y+2z\)

\(\Leftrightarrow\) 2\(\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\) \(\ge\) 2(x + y + z)

\(\Leftrightarrow\) \(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\ge x+y+z=10\)

Hay P_Min = 10 

Dấu "=" xảy ra \(\Leftrightarrow\) x = y = z = \(\dfrac{10}{3}\)

Vậy ...

Chúc bn học tốt!

Đặt \(A=\frac{x+y}{xyz}\)

Theo bài ra có ta có các số nguyên dương x,y,z có tổng =1

=> x+y+z=1

=> \(\left[\left(x+y\right)+z\right]^2=1\). Áp dụng BĐT \(\left(a+b\right)^2\ge4ab\)ta có:

\(1=\left[\left(x+y\right)+z\right]^2\ge4\left(x+y\right)z\)

Nhân 2 vế với số dương \(\frac{x+y}{xyz}\)được

\(\frac{x+y}{xyz}\ge\frac{4z\left(x+y\right)^2}{xyz}\ge\frac{4x\cdot4xy}{xyz}=16\)

Min_A=16 <=> \(\hept{\begin{cases}x+y=1\\x=y\\x+y+z=1\end{cases}\Leftrightarrow x=y=\frac{1}{4};z=\frac{1}{2}}\)

Vậy Min_A =16 đạt được khi \(x=y=\frac{1}{4};z=\frac{1}{2}\)

Ta có: \(1=x+y\ge2\sqrt{xy}\)

\(\Rightarrow4xy\le1\)

\(S=\frac{1}{x^2+y^2}+\frac{3}{4xy}\)

\(=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{4xy}\)

\(\ge\frac{4}{x^2+y^2+2xy}+\frac{1}{1}=\frac{4}{\left(x+y\right)^2}+1=\frac{4}{1}+1=5\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

Áp dụng BĐT AM - MG ta có :

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

Áp dụng BĐT Cauchy - Schwarz dạng Engel :

\(S\)\(=\)\(\frac{1}{x^2+y^2}\)\(-\)\(\frac{3}{4xy}\)\(=\)\(\frac{1}{x^2+y^2}\)\(-\)\(\frac{2}{4xy}\)\(-\)\(\frac{1}{4xy}\)

\(=\)\(\frac{1}{x^2+y^2}\)\(-\)\(\frac{1}{2xy}\)\(-\)\(\frac{1}{4xy}\)\(\ge\)\(\frac{\left(1-1\right)^2}{x^2-y^2-2xy}\)\(-\)\(\frac{1}{4xy}\)

\(\ge\)\(\frac{\left(1+1\right)^2}{\left(x+y\right)^2}\)\(-\)\(\frac{1}{4.\frac{1}{4}}\)\(=\)\(4\)\(-\)\(1\)\(=\)\(5\)

Xảy ra khi  \(x\)\(=\)\(y\)\(=\)\(\frac{1}{2}\)

Ta có: \(Q=\dfrac{2}{x^2+y^2}+\dfrac{3}{xy}=\dfrac{2}{x^2+y^2}+\dfrac{6}{2xy}=\dfrac{2}{x^2+y^2}+\dfrac{2}{2xy}+\dfrac{4}{2xy}\)

Áp dụng BĐT phụ: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

\(\Rightarrow2\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)\ge2\left(\dfrac{4}{x^2+2xy+y^2}\right)=2\left[\dfrac{4}{\left(x+y\right)^2}\right]=2.\dfrac{4}{4}=2\)

Dấu "=" xảy ra khi x=y=1

Áp dụng BĐT phụ: \(ab\le\dfrac{\left(a+b\right)^2}{4}\)

\(\Rightarrow xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{2^2}{4}=1\)

Dấu"=" xảy ra khi x=y=1

\(\Rightarrow2xy\le2.1=2\)

\(\Rightarrow\dfrac{4}{2xy}\ge\dfrac{4}{2}=2\)

\(\Rightarrow Q=\dfrac{2}{x^2+y^2}+\dfrac{2}{2xy}+\dfrac{4}{2xy}=\dfrac{2}{x^2+y^2}+\dfrac{3}{xy}\ge2+2=4\)

Dấu"=" xảy ra khi x=y=1

Lần sau tìm nơi gõ công thức và gõ hẳn ra nhé e <3

Áp dụng BĐT Cauchy-Schwarz ta có:

\(P=x^4+y^4\ge\frac{\left(x^2+y^2\right)^2}{2}\ge\frac{\left(\frac{\left(x+y\right)^2}{2}\right)^2}{2}=\frac{\left(\frac{2^2}{2}\right)^2}{2}=...\text{(tự tính nhé :)}\)

Khi \(x=y=1\)

I spring. Because spring has many beautiful  flowers.

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

Áp dụng BĐT Schwarz : \(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\ge\dfrac{\left(1+1\right)^2}{x^2+y^2+2xy}=\dfrac{4}{\left(x+y\right)^2}=4\)

Lại có \(\dfrac{1}{2xy}=\dfrac{2}{4xy}\ge\dfrac{2}{\left(x+y\right)^2}=2\)

Cộng vế với vế được P \(\ge6\) ("=" khi x = y = 1/2)

Vậy Min P = 6 <=> x = y = 1/2 

\(A\ge\dfrac{\left(1+2\right)^2}{x+y}=9\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(\dfrac{1}{3};\dfrac{2}{3}\right)\)

\(\frac{18}{x}+\frac{2}{y}=1\)

\(\Rightarrow\frac{1}{2}=\frac{9}{x}+\frac{1}{y}\)

\(\Rightarrow\frac{1}{2}=\frac{3^2}{x}+\frac{1}{2}\ge\frac{\left(3+1\right)^2}{x+y}\)

\(\Rightarrow\frac{1}{2}\ge\frac{16}{x+y}\)

\(\Rightarrow x+y\ge32\)

\(\text{Dấu '' = '' xảy ra khi:}\)

\(\orbr{\begin{cases}\frac{3}{x}=\frac{1}{y}\\x+y=32\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3y\\3y+y=32\end{cases}}\)          \(\Rightarrow\orbr{\begin{cases}x=24\\y=8\end{cases}}\)

đk : \(ĐK:x\ne0;y\ne0\)

Chia cả 2 vế cho 2, ta được: \(\frac{9}{x}+\frac{1}{y}=\frac{1}{2}\)

Áp dụng bất đẳng thức Svac-sơ : \(\frac{a^2}{b}+\frac{c^2}{d}\ge\frac{\left(a+c\right)^2}{b+d}\)

          \(\rightarrow VT\ge\frac{\left(3+1\right)^2}{x+y}\)\(\leftrightarrow\frac{1}{2}\ge\frac{\left(3+1\right)^2}{x+y}=\frac{16}{x+y}\)

                                  \(\Rightarrow x+y\ge32\)

                                  Dấu ''='' xảy ra \(\leftrightarrow\)\(\hept{\begin{cases}x=24\\y=8\end{cases}}\)

                             Vậy : \(Min\left(...\right)=32\leftrightarrow\hept{\begin{cases}x=24\\y=8\end{cases}}\)