Đặt: \(\frac{1}{y}=t\)> 0

Ta có: \(x+t\le1\)

\(P=\frac{xt}{2}+\frac{1}{xt}=\frac{xt}{2}+\frac{1}{32xt}+\frac{31}{32xt}\ge2\sqrt{\frac{xt}{2}.\frac{1}{32xt}}+\frac{31}{\frac{32\left(x+t\right)^2}{4}}=\frac{33}{8}\)

Dấu "=" xảy ra <=> x = t = 1/2 hay x = 1/2 và y = 2 

Vậy GTNN của P = 33/8 đạt tại x =1/2 và y =2 .

Ta có: \(1\ge x+y\ge2\sqrt{xy}\Rightarrow1\ge4xy\Rightarrow\frac{1}{xy}\ge4\)

\(\Rightarrow P\ge2\sqrt{\frac{1}{xy}}\cdot\sqrt{1+x^2y^2}=2\sqrt{\frac{1}{xy}+xy}\)

Mà \(\frac{1}{xy}+xy=\frac{15}{16}\cdot\frac{1}{xy}+\frac{1}{16xy}+xy\)

\(\ge\frac{15}{16}\cdot4+2\sqrt{\frac{1}{16xy}\cdot xy}=\frac{15}{16}\cdot4+\frac{2}{4}=\frac{17}{4}\)

\(\Rightarrow P\ge2\cdot\frac{\sqrt{17}}{2}=\sqrt{17}\) xảy ra khi \(x=y=\frac{1}{2}\)

v~ máy mk ko gõ dc chữ "x" 

cám ơn thắng nhìu

tự làm là hạnh phúc của mỗi công dân.

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 <=> \(x=y=\frac{1}{2}\)

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

Tương tự ta có:

\(\frac{y+1}{z^2+1}\ge y+1-\frac{yz+z}{2}\)

\(\frac{z+1}{1+x^2}\ge z+1-\frac{zx+x}{2}\)

Cộng vế theo vế ta có:

\(Q\ge3+\left(x+y+z\right)-\frac{x+y+z+xy+yz+zx}{2}\)

\(=3+\frac{x+y+z-xy-yz-zx}{2}\)

Có BĐT phụ sau:

\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\) ( tự cm )

\(\Rightarrow xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}=3\)

Khi đó \(P\ge3\)

Dấu "=" xảy ra tại \(x=y=z=1\)

Ta có: \(\frac{x+1}{y^2+1}=\left(x+1\right).\frac{1}{y^2+1}=\left(x+1\right)\left(1-\frac{y^2}{y^2+1}\right)\)

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

Thiết lập hai BĐT còn lại tương tự và cộng theo vế:

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

\(=6-\frac{\left(xy+yz+zx\right)+\left(x+y+z\right)}{2}\) (*)

Lại có BĐT \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)

Thật vậy,ta có: BĐT \(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca\ge3ab+3bc+3ca\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca\ge0\)

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)\ge0\)

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

Thay vào (*),ta có: \(P\ge6-\frac{\left(xy+yz+zx\right)+\left(x+y+z\right)}{2}\)

\(\ge6-\frac{\frac{\left(x+y+z\right)^2}{3}+3}{2}=6-\frac{3+3}{2}=3\)

Dấu "=" xảy ra \(\Leftrightarrow x^2=y^2=z^2=1\Leftrightarrow x=y=z=1\)

Vậy \(P_{min}=3\Leftrightarrow x=y=z=1\)

Bài t đúng 100% nhá,đứa nào tk sai t nhở? ngon vô làm lại=)

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

\(\ge3\sqrt[3]{\frac{1}{xyz\left(x+1\right)\left(y+1\right)\left(z+1\right)}}\)

Mà theo BĐT AM - GM ta có tiếp:

\(xyz\le\left(\frac{x+y+z}{3}\right)^3=1\)

\(\left(x+1\right)\left(y+1\right)\left(z+1\right)\le\left(\frac{x+y+z+3}{3}\right)^3=8\)

\(\Rightarrow P\le\frac{3}{2}\)

Đẳng thức xảy ra tại x=y=z=1

Vậy..................

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# Hok tốt !

\(x+y=1\Rightarrow\hept{\begin{cases}1-x=y\\1-y=x\end{cases}}\)

thay vào A ta được : \(A=\frac{1-y}{\sqrt{y}}+\frac{1-x}{\sqrt{x}}\)

\(\Rightarrow A=\frac{1}{\sqrt{y}}-\sqrt{y}+\frac{1}{\sqrt{x}}-\sqrt{x}\)

\(\Rightarrow A=\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)-\left(\sqrt{x}+\sqrt{y}\right)\)

áp dụng \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ta có : \(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\ge\frac{4}{\sqrt{x}+\sqrt{y}}\)

áp dụng \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\) ta có : \(\left(\sqrt{x}+\sqrt{y}\right)^2\le2\left(\sqrt{x}^2+\sqrt{y}^2\right)=2\)

\(\Rightarrow\sqrt{x}+\sqrt{y}\le\sqrt{2}\)

\(\Rightarrow A\ge\sqrt{8}-\sqrt{2}=\sqrt{2}\)

dấu = xảy ra khi a=y=1/2

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

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

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

Dấu "=" xảy ra khi x=y=1/2

Đúng ko biết !?