Chứng minh BĐT phụ:

\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)

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

Giờ thì chứng minh thôi:3

Áp dụng BĐT Cauchy-schwarz dạng engel ta có:

\(P=\left(2x+\frac{1}{x}\right)^2+\left(2y+\frac{1}{y}\right)^2\)

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

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

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

\(=8\)

Dấu "=" xảy ra khi và chỉ khi \(x=y=\frac{1}{2}\)

Vậy \(P_{min}=8\Leftrightarrow x=y=\frac{1}{2}\)

Bài này bạn làm đúng rồi nhưng mà bạn bị nhầm phép tính: \(\frac{\left[2\left(x+y\right)+\frac{4}{1}\right]^2}{2}=18\)

=> Min P=18

By Titu's Lemma we easy have:

\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

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

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

\(=\frac{17}{4}\)

Mk xin b2 nha!

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

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

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

\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)

Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)

Hãy xem phương pháp chọn điểm rơi của BĐT AM-GM( BĐT Cô-si)

Giải

\(P=\frac{3x}{10}+\frac{30}{x}+\frac{y}{20}+\frac{5}{y}+\frac{17x}{10}+\frac{19y}{20}\)

Áp dụng BĐT AM-GM, ta có:

\(\frac{3x}{10}+\frac{30}{x}\ge2\sqrt{\frac{3x}{10}\cdot\frac{30}{x}}=6\)

\(\frac{y}{20}+\frac{5}{y}\ge2\sqrt{\frac{y}{20}\cdot\frac{5}{y}}=1\)

Do đó

\(P\ge6+1+17+\frac{19}{2}=\frac{67}{2}\)(Vì \(x,y\ge10\))

Vậy \(P_{min}=\frac{67}{2}\Leftrightarrow x=y=10\)

<=> 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 \(x=y=5\)

Dự đoán dấu "=" khi x = 2 ; y= 1

Áp dụng bđt Cô-si cho 3 số và bđt \(\frac{a^2}{m}+\frac{b^2}{n}\ge\frac{\left(a+b\right)^2}{m+n}\) ta được

\(P=2x^2+y^2+\frac{28}{x}+\frac{1}{y}\)

    \(=\left(\frac{7x^2}{4}+\frac{14}{x}+\frac{14}{x}\right)+\left(\frac{y^2}{2}+\frac{1}{2y}+\frac{1}{2y}\right)+\left(\frac{x^2}{4}+\frac{y^2}{2}\right)\)

    \(\ge3\sqrt[3]{\frac{7x^2.14.14}{4.x^2}}+3\sqrt[3]{\frac{y^2.1.1}{2.2y.2y}}+\frac{\left(x+y\right)^2}{4+2}\)

      \(=3.\sqrt[3]{\frac{7.14.14}{4}}+\frac{3}{\sqrt[3]{2^3}}+\frac{3^2}{6}=24\)

Dấu "=" khi x = 2 ; y = 1 

Bài toán easy!

\(P=\left(2x^2+8\right)+\left(y^2+1\right)+\frac{28}{x}+\frac{1}{y}-9\)

Áp dụng BĐT AM-GM,ta có:

\(P\ge8x+2y+\frac{28}{x}+\frac{1}{y}-9\)

\(=\left(7x+\frac{28}{x}\right)+\left(y+\frac{1}{y}\right)+\left(x+y\right)-9\)

\(\ge2\sqrt{7x.\frac{28}{x}}+2\sqrt{y.\frac{1}{y}}+\left(x+y\right)-9\)

\(\ge28+2+3-9=24\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}2x^2=8\\y^2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)

Vậy \(P_{min}=24\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)