Đặt xy = a .

Ta có x + y = 1  => x^3 + y^3 = 1 - 3xy ( mũ 3 hai vế ) 

* Ta có a = xy \(\le\) \(\frac{\left(x+y\right)^2}{4}\) = \(\frac{1}{4}\) 
=> P = \(\frac{1}{1-3xy}\)+\(\frac{1}{xy}\)= \(\frac{1-2a}{a-3a^2}\). 

Để tìm min P thì ta tìm max \(\frac{1}{P}\)= Q <=> Q =  \(\frac{a-3a^2}{1-2a}\)

  Đặt A=(a-3a^2 )/(1-2a)
<=> A-2Aa=a-3a^2
<=> 3a^2 -a(1+2A)+A=0
Giải delta >=0 là 1 biểu thức theo A
từ đó tìm được min và max A

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: \(x=y=\frac{1}{2}\)

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

\(\Rightarrow\left(\sqrt{xy}+\frac{1}{4\sqrt{xy}}\right)^2+\frac{15}{16xy}+\frac{1}{2}\ge\frac{15}{16}\cdot4+\frac{1}{2}=\frac{17}{4}\)

Dấu "=" xảy ra <=> \(x=y=\frac{1}{2}\)

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

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

\(\ge\frac{4}{x^2+y^2+2xy}+2=\frac{4}{\left(x+y\right)^2}+2=6\)

Dấu "=" xảy ra khio x=y=1/2

Ta có : \(A=xy+\frac{1}{xy}=\left(16xy+\frac{1}{xy}\right)-15xy\)

Áp dụng bất đẳng thức Cauchy , ta có :

\(16xy+\frac{1}{xy}\ge2.\sqrt{16xy.\frac{1}{xy}}=8\)

Suy ra \(A\ge8-15xy\)

Ta lại có  \(xy\le\frac{\left(x+y\right)^2}{4}\)

<=> \(15xy\le\frac{15.1}{4}=\frac{15}{4}\)

<=> \(-15xy\ge\frac{15}{4}\)

Suy ra \(A\ge8-\frac{15}{4}=\frac{17}{4}\)

Đẳng thức xảy ra <=> x = y = \(\frac{1}{2}\)

b) Ta có \(A=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge\frac{\left(x+y+z\right)^2}{y+z+z+x+x+y}\)(BĐT Schwarz) 

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

Dấu "=" xảy ra khi \(\hept{\begin{cases}\frac{x^2}{y+z}=\frac{y^2}{z+x}=\frac{z^2}{x+y}\\x+y+z=2\end{cases}}\Leftrightarrow x=y=z=\frac{2}{3}\)

a) Có \(P=1.\sqrt{2x+yz}+1.\sqrt{2y+xz}+1.\sqrt{2z+xy}\)

\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(2x+yz+2y+xz+2z+xy\right)}\)(BĐT Bunyakovsky) 

\(=\sqrt{3.\left[2\left(x+y+z\right)+xy+yz+zx\right]}\)

\(\le\sqrt{3\left[4+\frac{\left(x+y+z\right)^2}{3}\right]}=\sqrt{3\left(4+\frac{4}{3}\right)}=4\)

Dấu "=" xảy ra <=> x = y = z = 2/3 

Đặt \(\hept{\begin{cases}\sqrt{2x+3}=a\left(a>0\right)\\\sqrt{y}=b\left(b\ge0\right)\end{cases}}\)

Thì ta có

\(\frac{b^2}{a^2}=\frac{a+1}{b+1}\)

\(\Leftrightarrow b^3+b^2=a^3+a^2\)

\(\Leftrightarrow\left(b-a\right)\left(b^2+ab+a^2\right)+\left(b-a\right)\left(b+a\right)=0\)

\(\Leftrightarrow\left(b-a\right)\left(b^2+ab+a^2+b+a\right)=0\)

Mà \(\left(b^2+ab+a^2+b+a\right)>0\)

\(\Rightarrow a=b\)

\(\Rightarrow2x+3=y\)

Thế vào Q ta được 

\(Q=2x^2-5x-12=\left(2x^2-\frac{2x\times\sqrt{2}\times5}{2\sqrt{2}}+\frac{25}{8}\right)-\frac{121}{8}\)

\(=\left(\sqrt{2}x-\frac{5}{2\sqrt{2}}\right)^2-\frac{121}{8}\ge\frac{-121}{8}\)

1. \(1=x^2+y^2\ge2xy\Rightarrow xy\le\frac{1}{2}\)

 \(A=-2+\frac{2}{1+xy}\ge-2+\frac{2}{1+\frac{1}{2}}=-\frac{2}{3}\)

max A = -2/3 khi x=y=\(\frac{\sqrt{2}}{2}\)

\(\frac{1}{xy}+\frac{1}{xz}=\frac{1}{x}\left(\frac{1}{y}+\frac{1}{z}\right)\ge\frac{1}{x}.\frac{4}{y+z}=\frac{4}{\left(4-t\right)t}=\frac{4}{4-\left(t-2\right)^2}\ge1\) với t = y+z => x =4 -t

\(A=x^2+y^2=\frac{\left(1^2+1^2\right)\left(x^2+y^2\right)}{2}\ge\frac{\left(1.x+1.y\right)^2}{2}=\frac{1}{2}\)A min = 1 khi x =y = 1/2

\(\sqrt{A}=\sqrt{x^2+y^2}\le\sqrt{x^2}+\sqrt{y^2}=x+y=1\)( \(\sqrt{a+b}\le\sqrt{a}+\sqrt{b}\))

=> A\(\le1\) => Max A = 1 khi x =0;y =1 hoặc x =1 ; y =0

???????

\(A=\frac{1}{x^2+y^2}+\frac{5}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{9}{2xy}\ge\frac{4}{\left(x+y\right)^2}+\frac{9}{2\left(\frac{x+y}{2}\right)^2}\)

nên  \(A\ge4+9.2=22\)

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