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\(Gt\Rightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\)
Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\Rightarrow ab+bc+ca=1\)
\(VT=\frac{2}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}+\frac{1}{\sqrt{1+z^2}}\)
\(=\frac{\frac{2}{x}}{\sqrt{\frac{1}{x^2}+1}}+\frac{\frac{1}{y}}{\sqrt{\frac{1}{y^2}+1}}+\frac{\frac{1}{z}}{\sqrt{\frac{1}{z^2}+1}}\)
\(=\frac{2a}{\sqrt{a^2+ab+bc+ca}}+\frac{b}{\sqrt{b^2+ab+bc+ca}}+\frac{c}{\sqrt{c^2+ab+bc+ca}}\)
\(=\sqrt{\frac{2a}{\left(a+b\right)}\cdot\frac{2a}{\left(a+c\right)}}+\sqrt{\frac{2b}{\left(b+a\right)}\cdot\frac{b}{2\left(b+c\right)}}\)\(+\sqrt{\frac{2c}{\left(c+a\right)}\cdot\frac{c}{2\left(c+b\right)}}\)
\(\le\frac{\frac{2a}{a+b}+\frac{2a}{a+c}+\frac{2b}{a+b}+\frac{b}{2\left(b+c\right)}+\frac{2c}{c+a}+\frac{c}{2\left(c+b\right)}}{2}=\frac{9}{4}\)
Bài 2:
\(a^4+b^4\ge a^3b+b^3a\)
\(\Leftrightarrow a^4-a^3b+b^4-b^3a\ge0\)
\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)
ta thấy : \(\orbr{\orbr{\begin{cases}\left(a-b\right)^2\ge0\\\left(a^2+ab+b^2\right)\ge0\end{cases}}}\Leftrightarrow dpcm\)
Dấu " = " xảy ra khi a = b
tk nka !!!! mk cố giải mấy bài nữa !11
\(VT\le\sqrt{2\left(1+2x+1+2y\right)}=2\sqrt{1+x+y}\)
\(VT\le2\sqrt{1+\sqrt{2\left(x^2+y^2\right)}}=2\sqrt{3}\)
Dấu "=" xảy ra khi \(x=y=1\)
a) Có \(\left(x-y\right)^2\ge0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-y\right)^2\ge\left(x+y\right)^2\)
\(\Leftrightarrow x^2+2xy+y^2+x^2-2xy+y^2\ge\left(x+y\right)^2\)
\(\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)
\(\Leftrightarrow2\ge\left(x+y\right)^2\)
\(\Leftrightarrow\left|x+y\right|\le\sqrt{2}\)
\(\Leftrightarrow-\sqrt{2}\le x+y\le\sqrt{2}\)( đpcm )
Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{\pm\sqrt{2}}{2}\)
b) Áp dụng bđt Cô-si :
\(\frac{1}{x}+\frac{1}{y}\ge2\sqrt{\frac{1}{xy}}=\frac{2}{\sqrt{xy}}\)
Chứng minh tương tự rồi cộng vế ta có :
\(2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge2\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\right)\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\)( đpcm )
Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)
a) Theo BĐT Bunhiacopxki suy ra \(2=2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)
Do đó suy ra \(-\sqrt{2}\le x+y\le\sqrt{2}\)
b) Đặt \(\frac{1}{\sqrt{x}}=a;\frac{1}{\sqrt{y}}=b;\frac{1}{\sqrt{z}}=c\)
Cần chứng minh \(a^2+b^2+c^2\ge ab+bc+ca\Leftrightarrow\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{2}\ge0\) (đúng)
Xảy ra đẳng thức khi a = b = c hay x = y = z
Đề phải cho \(x,y\) dương nữa!
Giải:
Ta có: \(xy\left(x+y\right)^2\le\dfrac{1}{64}\)
\(\Leftrightarrow\sqrt{xy\left(x+y\right)^2}\le\sqrt{\dfrac{1}{64}}\)
\(\Leftrightarrow\sqrt{xy}\left(x+y\right)\le\dfrac{1}{8}\)
Vậy ta cần chứng minh BĐT tương đương \(\sqrt{xy}\left(x+y\right)\le\dfrac{1}{8}\)
Áp dụng BĐT AM - GM ta có:
\(\sqrt{xy}\left(x+y\right)=\dfrac{1}{2}.2\sqrt{xy}\left(x+y\right)\)
\(\le\dfrac{1}{2}.\dfrac{x+y+2\sqrt{xy}}{4}=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^4}{8}\) \(=\dfrac{1}{8}\)
\(\Rightarrow xy\left(x+y\right)^2\le\dfrac{1}{64}\) (Đpcm)
Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{4}\)
\(2\le\sqrt{x}+\sqrt{4-x}\le2\sqrt{2}\) (1) (ĐK: \(\left\{{}\begin{matrix}x\ge0\\4-x\ge0\end{matrix}\right.\)\(\Leftrightarrow0\le x\le4\))
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}2\le\sqrt{x}+\sqrt{4-x}\\\sqrt{x}+\sqrt{4-x}\le2\sqrt{2}\end{matrix}\right.\) (\(0\le x\le4\))
\(\Leftrightarrow\left\{{}\begin{matrix}4\le4+2\sqrt{x\left(4-x\right)}\\4+2\sqrt{x\left(4-x\right)}\le8\end{matrix}\right.\) (\(0\le x\le4\))
\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x\left(4-x\right)}\ge0\\\sqrt{x\left(4-x\right)}\le2\end{matrix}\right.\)(\(0\le x\le4\))
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(4-x\right)\le4\\0\le x\le4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\0\le x\le4\end{matrix}\right.\) (đpcm)