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Lời giải;
Vế 1:
Áp dụng BĐT AM-GM:
$2=(x^2+y^2)(1+1)\geq (x+y)^2\Rightarrow x+y\leq \sqrt{2}$
$x^3+\frac{x}{2}\geq \sqrt{2}x^2$
$y^3+\frac{y}{2}\geq \sqrt{2}y^2$
$\Rightarrow x^3+y^3+\frac{x+y}{2}\geq \sqrt{2}(x^2+y^2)=\sqrt{2}$
$\Rightarrow x^3+y^3\geq \sqrt{2}-\frac{x+y}{2}\geq \sqrt{2}-\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$
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Vế 2:
$x^2+y^2=1$
$\Rightarrow x^2=1-y^2\leq 1\Rightarrow -1\leq x\leq 1$
$y^2=1-x^2\leq 1\Rightarrow -1\leq y\leq 1$
$\Rightarrow x^3\leq x^2; y^3\leq y^2$
$\Rightarrow x^3+y^3\leq x^2+y^2$ hay $x^3+y^3\leq 1$
\(\frac{x\sqrt{y}+y\sqrt{x}}{x+y}-\frac{x+y}{2}\le\frac{x\sqrt{y}+y\sqrt{x}}{2\sqrt{xy}}-\frac{x+y}{2}=\frac{\sqrt{x}+\sqrt{y}}{2}-\frac{x+y}{2}\)
Cần chứng minh : \(\frac{\sqrt{x}+\sqrt{y}}{2}-\frac{x+y}{2}\le\frac{1}{4}\Leftrightarrow\sqrt{x}+\sqrt{y}-x-y\le\frac{1}{2}\Leftrightarrow2\sqrt{x}+2\sqrt{y}-2x-2y\le1\)
\(\Leftrightarrow2x+2y-2\sqrt{x}-2\sqrt{y}+1\ge0\)\(\Leftrightarrow\left(\sqrt{2x}-\frac{1}{\sqrt{2}}\right)^2+\left(\sqrt{2y}-\frac{1}{\sqrt{2}}\right)^2\ge0\)
Vì BĐT cuối luôn đúng nên BĐT cần chứng minh luôn đúng khi x = y = \(\frac{1}{4}\)
\(VT=\frac{x\sqrt{y}+y\sqrt{x}}{x+y}-\frac{x+y}{2}\le\frac{\sqrt{2xy\left(x+y\right)}}{x+y}-\frac{x+y}{2}\)
\(\le\frac{\left(x+y\right)\sqrt{\frac{x+y}{2}}}{x+y}-\frac{x+y}{2}\) . Cm : \(\sqrt{\frac{x+y}{2}}-\frac{x+y}{2}\le\frac{1}{4}\)
Đặt \(x+y=t>0\)thì :
\(\sqrt{\frac{t}{2}}-\frac{t}{2}\le\frac{1}{4}\Leftrightarrow-\frac{1}{4}\left(\sqrt{2t}-1\right)^2\le0\) ( đúng )
Chúc bạn học tốt !!!
c1: phân tích từng cái
c2, nhân x cho (1) y cho 2
sau đs dùng bunhia
từ x+y=1
=> x^2-xy+y^2...
A(BT)=1/9((9/x+y+1) +(9/y+z+1)+9/(z+x+1)<=1/9(1/x+1/y+1+1/y+1/z+1+1/z+1/x+1)=1/9(2/x+2/y+2/z+3)
=1/9(2.(xy+yz+zx)/xyz)+3=2/9(xy+yz+zx)+1/3<=2/9.3+1/3=1(đpcm)
Another way :|
Đặt \(\hept{\begin{cases}a=\sqrt[3]{x}\\b=\sqrt[3]{y}\\c=\sqrt[3]{z}\end{cases}}\Rightarrow\hept{\begin{cases}x=a^3\\y=b^3\\z=c^3\end{cases}}\)và \(xyz=1\Rightarrow\left(abc\right)^3=1\Rightarrow abc=1\)
Áp dụng BĐT AM-GM ta có:\(a^3+b^3+1=a^3+b^3+abc\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+abc\)
\(\ge ab\left(a+b\right)+abc=ab\left(a+b+c\right)\)
\(\Rightarrow\frac{1}{a^3+b^3+1}\le\frac{1}{ab\left(a+b+c\right)}\). Tương tự cũng có:
\(\frac{1}{b^3+c^3+1}\le\frac{1}{bc\left(a+b+c\right)};\frac{1}{c^3+a^3+1}\le\frac{1}{ca\left(a+b+c\right)}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le\frac{1}{ab\left(a+b+c\right)}+\frac{1}{bc\left(a+b+c\right)}+\frac{1}{ca\left(a+b+c\right)}\)
\(=\frac{c}{abc\left(a+b+c\right)}+\frac{a}{abc\left(a+b+c\right)}+\frac{b}{abc\left(a+b+c\right)}=\frac{a+b+c}{abc\left(a+b+c\right)}=1\)
Xảy ra khi \(a=b=c=1\Rightarrow x=y=z=1\)
\(\frac{x\sqrt{y}+y\sqrt{x}}{x+y}-\frac{x+y}{2}\le\frac{1}{4}\)
Ta có:
\(VT\le\frac{x\sqrt{y}+y\sqrt{x}}{2\sqrt{xy}}-\frac{x+y}{2}\)
\(=\frac{\sqrt{x}+\sqrt{y}}{2}-\frac{x+y}{2}\)
Giờ ta chỉ cần chứng minh
\(\frac{\sqrt{x}+\sqrt{y}}{2}-\frac{x+y}{2}\le\frac{1}{4}\)
\(\Leftrightarrow2x+2y-2\sqrt{x}-2\sqrt{y}+1\ge0\)
\(\Leftrightarrow\left(2x-2\sqrt{x}+\frac{1}{2}\right)+\left(2y-2\sqrt{y}+\frac{1}{2}\right)\ge0\)
\(\Leftrightarrow\left(\sqrt{2x}-\frac{1}{\sqrt{2}}\right)^2+\left(\sqrt{2y}-\frac{1}{\sqrt{2}}\right)^2\ge0\)(đúng)
Dấu = xảy ra khi \(x=y=\frac{1}{4}\)