# Bài 1

* Ta cm BĐT sau \(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\) (1) bằng cách biến đổi tương đương

* Với \(x,y>0\) áp dụng (1) ta có

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

Mà \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2}\)

\(\Rightarrow\) \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)^2\le1\) \(\Leftrightarrow\) \(0< \dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\le1\) (I)

* Ta cm BĐT phụ \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) với \(a,b>0\) (2)

Áp dụng (2) với x , y > 0 ta có

\(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\ge\dfrac{4}{\sqrt{x}+\sqrt{y}}\) (II)

* Từ (I) và (II) \(\Rightarrow\) \(\dfrac{4}{\sqrt{x}+\sqrt{y}}\le1\)

\(\Leftrightarrow\) \(\sqrt{x}+\sqrt{y}\ge4\)

Dấu "=" xra khi \(x=y=4\)

Vậy min \(\sqrt{x}+\sqrt{y}=4\) khi \(x=y=4\)

Bài 1 :

Ta có : \(\dfrac{1}{3a^2+b^2}+\dfrac{2}{b^2+3ab}=\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\)

Theo BĐT Cô - Si dưới dạng engel ta có :

\(\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\ge\dfrac{\left(1+2\right)^2}{3a^2+6ab+3b^2}=\dfrac{9}{3\left(a+b\right)^2}=\dfrac{9}{3.1}=3\)

Dấu \("="\) xảy ra khi : \(a=b=\dfrac{1}{2}\)

\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{2xy}\ge\dfrac{4}{\left(x+y\right)^2}+\dfrac{1}{2xy}\ge\dfrac{4}{1^2}+\dfrac{1}{\dfrac{2.\left(x+y\right)^2}{4}}\ge4+2=6\)

Dấu "=" xảy ra <=> x = y = 0,5

áp dụng BDT AM-GM \(=>x+y\ge2\sqrt{xy}=>\left(x+y\right)^2\ge4xy\left(1\right)\)

mà \(x+y\le1=>\left(x+y\right)^2\le1\left(2\right)\)

(1)(2)\(=>4xy\le\left(x+y\right)^2\le1=>4xy\le1=>xy\le\dfrac{1}{4}\)

\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\sqrt{1+x^2y^2}\ge2\sqrt{\dfrac{1+x^2y^2}{xy}}=2\sqrt{\dfrac{1}{xy}+xy}\)

\(=2\sqrt{\dfrac{1}{xy}+16xy-15xy}=2\sqrt{2\sqrt{16}-\dfrac{15}{4}}=\sqrt{17}\)

dấu"=" xảy ra<=>\(x=y=\dfrac{1}{2}\)

\(1\ge x+y\ge2\sqrt{xy}\Rightarrow xy\le\dfrac{1}{4}\Rightarrow\dfrac{1}{xy}\ge4\)

Ta có:

\(A\ge\dfrac{2}{\sqrt{xy}}.\sqrt{1+x^2y^2}=2\sqrt{\dfrac{1}{xy}+xy}=2\sqrt{\left(xy+\dfrac{1}{16xy}\right)+\dfrac{15}{16}.\dfrac{1}{xy}}\)

\(A\ge2\sqrt{2\sqrt{\dfrac{xy}{16xy}}+\dfrac{15}{16}.4}=\sqrt{17}\)

\(A_{min}=\sqrt{17}\) khi \(x=y=\dfrac{1}{2}\)

\(c,P=\dfrac{x^2-x^2+8xy-16y^2}{x^2+4y^2}=\dfrac{8\left(\dfrac{x}{y}\right)-16}{\left(\dfrac{x}{y}\right)^2+4}\)

Đặt \(\dfrac{x}{y}=t\)

\(\Leftrightarrow P=\dfrac{8t-16}{t^2+4}\Leftrightarrow Pt^2+4P=8t-16\\ \Leftrightarrow Pt^2-8t+4P+16=0\)

Với \(P=0\Leftrightarrow t=2\)

Với \(P\ne0\Leftrightarrow\Delta'=16-P\left(4P+16\right)\ge0\)

\(\Leftrightarrow-P^2-4P+4\ge0\Leftrightarrow-2-2\sqrt{2}\le P\le-2+2\sqrt{2}\)

Vậy \(P_{max}=-2+2\sqrt{2}\Leftrightarrow t=\dfrac{4}{P}=\dfrac{4}{-2+2\sqrt{2}}=2+\sqrt{2}\)

\(\Leftrightarrow\dfrac{x}{y}=2+2\sqrt{2}\)

Bài a hình như sai đề rồi bạn.

\(a,\text{Đặt }\left\{{}\begin{matrix}S=y+z\\P=yz\end{matrix}\right.\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\left(y+z\right)^2-2yz+x^2=8\\x\left(y+z\right)+yz=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}S^2-2P+x^2=8\\Sx+P=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}S^2-2\left(4-Sx\right)+x^2=8\\P=4-Sx\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}S^2+2Sx+x^2-16=0\left(1\right)\\P=4-Sx\end{matrix}\right.\\ \left(1\right)\Leftrightarrow\left(S+x-4\right)\left(S+x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}S=-x+4\Rightarrow P=\left(x-2\right)^2\\S=-x-4\Rightarrow P=\left(x+2\right)^2\end{matrix}\right.\)

Mà y,z là nghiệm của hệ nên \(S^2-4P\ge0\Leftrightarrow\left[{}\begin{matrix}\left(4-x\right)^2\ge4\left(x-2\right)^2\\\left(-4-x\right)^2\ge4\left(x+2\right)^2\end{matrix}\right.\Leftrightarrow-\dfrac{8}{3}\le x\le\dfrac{8}{3}\)

\(1,\dfrac{1}{1+x}=1-\dfrac{1}{1+y}+1-\dfrac{1}{1+z}=\dfrac{y}{1+y}+\dfrac{z}{1+z}\ge2\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}\)

Cmtt: \(\dfrac{1}{1+y}\ge2\sqrt{\dfrac{xz}{\left(1+x\right)\left(1+z\right)}};\dfrac{1}{1+z}\ge2\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}\)

Nhân VTV

\(\Leftrightarrow\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge8\sqrt{\dfrac{x^2y^2z^2}{\left(1+x\right)^2\left(1+y\right)^2\left(1+z\right)^2}}\\ \Leftrightarrow\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge\dfrac{8xyz}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\\ \Leftrightarrow8xyz\le1\Leftrightarrow xyz\le\dfrac{1}{8}\)

Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{2}\)

\(2,\\ a,2x^2+y^2-2xy=1\\ \Leftrightarrow\left(x-y\right)^2+x^2=1\\ \Leftrightarrow\left(x-y\right)^2=1-x^2\ge0\\ \Leftrightarrow x^2\le1\Leftrightarrow\sqrt{x^2}\le1\Leftrightarrow\left|x\right|\le1\)