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C.hóa \(x+y=1\) và dùng C-S:
\(VT^2\le\frac{2x}{\left(y+1\right)^2}+\frac{2y}{\left(x+1\right)^2}\le\frac{8}{9}=VP^2\)
\(BDT\Leftrightarrow\frac{x}{\left(2-x\right)^2}+\frac{y}{\left(2-y\right)^2}\le\frac{4}{9}\left(1\right)\)
Ta có BĐT phụ \(\frac{x}{\left(2-x\right)^2}\le\frac{20}{27}x-\frac{4}{27}\)
\(\Leftrightarrow-\frac{\left(2x-1\right)^2\left(5x-16\right)}{27\left(x-2\right)^2}\le0\) *Đúng*
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT_{\left(1\right)}\le\frac{20}{27}\left(x+y\right)-\frac{4}{27}\cdot2=\frac{4}{9}=VP_{\left(1\right)}\)
"=" khi \(x=y=\frac{1}{2}\)
Ta có : \(xy\left(x+y\right)^2\le\frac{1}{64}\)\(\Rightarrow\)\(\sqrt{xy\left(x+y\right)^2}\le\sqrt{\frac{1}{64}}\)
\(\Rightarrow\)\(\sqrt{xy}\left(x+y\right)\le\frac{1}{8}\)
ta cần c/m \(\sqrt{xy}\left(x+y\right)\le\frac{1}{8}\)
Thật vậy, ta có
Áp dụng BĐT : \(ab\le\frac{\left(a+b\right)^2}{4}\). Dấu "=" xảy ra \(\Leftrightarrow\)a = b
\(\sqrt{xy}\left(x+y\right)=\frac{1}{2}.2\sqrt{xy}\left(x+y\right)\le\frac{1}{2}.\frac{\left(x+2\sqrt{xy}+y\right)^2}{4}=\frac{\left(\sqrt{x}^2+2\sqrt{xy}+\sqrt{y}^2\right)^2}{4}.\frac{1}{2}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^4}{8}=\frac{1}{8}\)
Dấu " = " xảy ra \(\Leftrightarrow\)\(x=y=\frac{1}{4}\)
\(\text{Ta có:}\frac{x}{y+1}+\frac{y}{x+1}=\frac{x^2+x+y^2+y}{\left(x+1\right)\left(y+1\right)}\)
\(=\frac{\left(x+y\right)^2-2xy+1}{xy+x+y+1}=\frac{1-2xy+1}{xy+2}\)
\(=\frac{2-2xy}{2+xy}\)
\(\text{Vì }2-2xy\le2+xy\left(do\text{ x,y không âm}\right)\text{ nên }\frac{2-2xy}{2+xy}\le1\)
\(=>\frac{x}{y+1}+\frac{y}{x+1}\le1\)
Sửa đề nhé\(\dfrac{1}{3x+3y+2z}=\dfrac{1}{\left(z+x\right)+\left(z+y\right)+\left(x+y\right)+\left(x+y\right)}\)
\(\le\dfrac{1}{16}\left(\dfrac{1}{x+z}+\dfrac{1}{z+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}\right)\)
CMTT và cộng theo vế:
\(VT\le\dfrac{1}{16}\left(\dfrac{1}{x+z}+\dfrac{1}{z+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}+\dfrac{1}{x+z}+\dfrac{1}{x+z}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{y+z}\right)\)
\(=\dfrac{1}{16}.24=\dfrac{3}{2}\)
\("="\Leftrightarrow x=y=z=\dfrac{1}{4}\)
Do \(\left\{{}\begin{matrix}x;y\ge0\\x+y=1\end{matrix}\right.\) \(\Rightarrow0\le x;y\le1\) \(\Rightarrow\left\{{}\begin{matrix}x^2\le x\\y^2\le y\end{matrix}\right.\)
\(\Rightarrow x^2+y^2\le x+y=1\)
\(P=\dfrac{x}{y+1}+\dfrac{y}{x+1}=\dfrac{x^2+y^2+x+y}{\left(x+1\right)\left(y+1\right)}=\dfrac{x^2+y^2+1}{xy+x+y+1}\)
\(=\dfrac{x^2+y^2+1}{xy+2}\le\dfrac{x^2+y^2+1}{2}\le\dfrac{1+1}{2}=1\) (đpcm)
Dấu "=" xảy ra khi \(\left(x;y\right)=\left(1;0\right);\left(0;1\right)\)