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Áp dụng bất đẳng thức cauchy:
\(P=\sum\dfrac{x^2\left(y+z\right)}{y\sqrt{y}+2z\sqrt{z}}\ge\sum\dfrac{2x^2\sqrt{yz}}{y\sqrt{y}+2z\sqrt{z}}=\sum\dfrac{2\sqrt{x^3}\sqrt{xyz}}{\sqrt{y^3}+2\sqrt{z^3}}=\sum\dfrac{2\sqrt{x^3}}{\sqrt{y^3}+2\sqrt{z^3}}\)(vì xyz=1).
đặt \(\left\{{}\begin{matrix}\sqrt{x^3}=a\\\sqrt{y^3}=b\\\sqrt{z^3}=c\end{matrix}\right.\)(\(a,b,c>0\))thì giả thiết trở thành cho abc=1. tìm Min \(P=\dfrac{2a}{b+2c}+\dfrac{2b}{c+2a}+\dfrac{2c}{a+2b}\)
Áp dụng BĐT cauchy-schwarz:
\(P=2\left(\dfrac{a^2}{ab+2ac}+\dfrac{b^2}{bc+2ab}+\dfrac{c^2}{ac+2bc}\right)\ge\dfrac{2\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=2\)( AM-GM \(3\left(ab+bc+ca\right)\le\left(a+b+c\right)^2\))
Dấu = xảy ra khi a=b=c=1 hay x=y=z=1
1. Với mọi số thực x;y;z ta có:
\(x^2+y^2+z^2+\dfrac{1}{2}\left(x^2+1\right)+\dfrac{1}{2}\left(y^2+1\right)+\dfrac{1}{2}\left(z^2+1\right)\ge xy+yz+zx+x+y+z\)
\(\Leftrightarrow\dfrac{3}{2}P+\dfrac{3}{2}\ge6\)
\(\Rightarrow P\ge3\)
\(P_{min}=3\) khi \(x=y=z=1\)
1.1
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}=a>0\\\dfrac{1}{\sqrt{y}}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+\sqrt{2-b^2}=2\\b+\sqrt{2-a^2}=2\end{matrix}\right.\)
\(\Rightarrow a-b+\sqrt{2-b^2}-\sqrt{2-a^2}=0\)
\(\Leftrightarrow a-b+\dfrac{\left(a-b\right)\left(a+b\right)}{\sqrt{2-b^2}+\sqrt{2-a^2}}=0\)
\(\Leftrightarrow a=b\Leftrightarrow x=y\)
Thay vào pt đầu:
\(a+\sqrt{2-a^2}=2\Rightarrow\sqrt{2-a^2}=2-a\) (\(a\le2\))
\(\Leftrightarrow2-a^2=4-4a+a^2\Leftrightarrow2a^2-4a+2=0\)
\(\Rightarrow a=1\Rightarrow x=y=1\)
2.
\(\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\x^2-xy+y^2=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x^2+3xy+3y^2=21\\7x^2-7xy+7y^2=21\end{matrix}\right.\)
\(\Rightarrow4x^2-10xy+4y^2=0\)
\(\Leftrightarrow2\left(2x-y\right)\left(x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2x\\y=\dfrac{1}{2}x\end{matrix}\right.\)
Thế vào pt đầu
...
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 \(x=y=\dfrac{1}{2}\)
\(a,\dfrac{x^2+x+2}{\sqrt{x^2+x+1}}=\dfrac{x^2+x+1+1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}+\dfrac{1}{\sqrt{x^2+x+1}}\left(1\right)\)
Áp dụng BĐT cosi: \(\left(1\right)\ge2\sqrt{\sqrt{x^2+x+1}\cdot\dfrac{1}{\sqrt{x^2+x+1}}}=2\)
Dấu \("="\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
\(B=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\sqrt{1+x^2y^2}\)
\(\ge\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\dfrac{\left(1+0,25xy\right)}{\sqrt{1,0625}}\)
\(=\dfrac{1}{\sqrt{1,0625}}\left(\dfrac{1}{x}+\dfrac{1}{y}+0,25x+0,25y\right)\)
\(=\dfrac{1}{\sqrt{1,0625}}\left(\left(\dfrac{1}{x}+4x\right)+\left(\dfrac{1}{y}+4y\right)-\dfrac{15}{4}\left(x+y\right)\right)\)
\(\ge\dfrac{1}{\sqrt{1,0625}}\left(4+4-\dfrac{15}{4}\right)=\sqrt{17}\)
bạn giải thích giùm mình dòng thứ 2 đi :) dùng bđt j v :)