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\(VT\le\dfrac{x}{2x+2y+2}+\dfrac{y}{2yz+2z+2}+\dfrac{z}{2z+2x+2}\)
Nên ta chỉ cần chứng minh: \(\dfrac{x}{x+y+1}+\dfrac{y}{y+z+1}+\dfrac{z}{z+x+1}\le1\)
\(\Leftrightarrow\dfrac{y+1}{x+y+1}+\dfrac{z+1}{y+z+1}+\dfrac{x+1}{z+x+1}\ge2\)
Thật vậy, ta có:
\(VT=\dfrac{\left(x+1\right)^2}{\left(x+1\right)\left(z+x+1\right)}+\dfrac{\left(y+1\right)^2}{\left(y+1\right)\left(x+y+1\right)}+\dfrac{\left(z+1\right)^2}{\left(z+1\right)\left(y+z+1\right)}\)
\(VT\ge\dfrac{\left(x+y+z+3\right)^2}{\left(x^2+y^2+z^2\right)+3\left(x+y+z\right)+xy+yz+zx+3}\)
\(VT\ge\dfrac{6\left(x+y+z\right)+2\left(xy+yz+zx\right)+12}{3\left(x+y+z\right)+xy+yz+zx+6}=2\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
Bài 1:
\((x,y,z)=(\frac{2a^2}{bc}; \frac{2b^2}{ca}; \frac{2c^2}{ab})\) (\(a,b,c>0\) )
Khi đó:
\(\text{VT}=\frac{\frac{4a^4}{b^2c^2}}{\frac{4a^4}{b^2c^2}+\frac{4a^2}{bc}+1}+\frac{\frac{4b^4}{c^2a^2}}{\frac{4b^4}{c^2a^2}+\frac{4b^2}{ca}+4}+\frac{\frac{4c^4}{a^2b^2}}{\frac{4c^4}{a^2b^2}+\frac{4c^2}{ab}+4}\)
\(=\frac{a^4}{a^4+a^2bc+b^2c^2}+\frac{b^4}{b^4+b^2ac+a^2c^2}+\frac{c^4}{c^4+c^2ab+a^2b^2}\)
\(\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+a^2bc+b^2ac+c^2ab+(a^2b^2+b^2c^2+c^2a^2)}\)
(Áp dụng BĐT Cauchy_Schwarz)
Theo BĐT Cauchy dễ thấy:
\(a^2b^2+b^2c^2+c^2a^2\geq a^2bc+b^2ca+c^2ab\)
\(\Rightarrow \text{VT}\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)}=\frac{(a^2+b^2+c^2)^2}{(a^2+b^2+c^2)^2}=1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=2$
Bài 2:
Đặt \((x,y,z)=\left(\frac{a}{b};\frac{b}{c}; \frac{c}{a}\right)\)
Ta có:
\(\text{VT}=\left(\frac{a}{b}+\frac{c}{b}-1\right)\left(\frac{b}{c}+\frac{a}{c}-1\right)\left(\frac{c}{a}+\frac{b}{a}-1\right)\)
\(=\frac{(a+c-b)(b+a-c)(c+b-a)}{abc}\)
Áp dụng BĐT Cauchy:
\((a+c-b)(b+a-c)\leq \left(\frac{a+c-b+b+a-c}{2}\right)^2=a^2\)
\((b+a-c)(c+b-a)\leq \left(\frac{b+a-c+c+b-a}{2}\right)^2=b^2\)
\((a+c-b)(c+b-a)\leq \left(\frac{a+c-b+c+b-a}{2}\right)^2=c^2\)
Nhân theo vế:
\(\Rightarrow [(a+c-b)(b+a-c)(c+b-a)]^2\leq (abc)^2\)
\(\Rightarrow (a+c-b)(b+a-c)(c+b-a)\leq abc\)
\(\Rightarrow \text{VT}\leq 1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=1$
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:
\(VT\le\dfrac{1}{16}\left(\dfrac{1}{x+z}+\dfrac{1}{z+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}+\dfrac{1}{x+z}+\dfrac{1}{x+z}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{y+z}\right)\)
\(=\dfrac{1}{16}.24=\dfrac{3}{2}\)
\("="\Leftrightarrow x=y=z=\dfrac{1}{4}\)
\(VT^2\le3\left(\dfrac{1}{2x^2+y^2+3}+\dfrac{1}{2y^2+z^2+3}+\dfrac{1}{2z^2+x^2+3}\right)\)
Mặt khác:
\(\dfrac{1}{2\left(x^2+1\right)+y^2+1}\le\dfrac{1}{4x+2y}=\dfrac{1}{2}\left(\dfrac{1}{x+x+y}\right)\le\dfrac{1}{18}\left(\dfrac{2}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow VT^2\le\dfrac{1}{6}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{3}{2}\)
\(\Rightarrow VT\le\dfrac{\sqrt{6}}{2}\)
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 : \(a=b=\dfrac{1}{2}\)
Ta có:\(\frac{4+4\sqrt{1+x^2}}{4x}\le\frac{4+5+x^2}{4x}=\)\(\frac{x^2+9}{4x}\)Tương tự ta đc P\(\le\frac{x+y+z}{4}+\frac{9}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(=\frac{1}{4}\left(x+y+z\right)+\frac{9}{4}\left(\frac{xy+yz+zx}{xyz}\right)\)\(\le\frac{1}{4}\left(x+y+z\right)+\frac{9}{4}\cdot\frac{\left(x+y+z\right)^2}{3\left(x+y+z\right)}\)\(=x+y+z\)
Dấu '='xảy ra <=>\(\hept{\begin{cases}x+y+z=xyz\\x=y=z\end{cases}\Rightarrow x=y=z=}\)\(\frac{1}{\sqrt{3}}\)
đặt\(A=\dfrac{x^3}{2x+3y+5z}+\dfrac{y^3}{2y+3z+5x}+\dfrac{z^3}{2z+3x+5y}\)
\(=>A=\dfrac{x^4}{2x^2+3xy+5xz}+\dfrac{y^4}{2y^2+3yz+5xy}+\dfrac{z^4}{2z^2+3xz+5yz}\)
BBDT AM-GM
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)
theo BDT AM -GM ta chứng minh được \(xy+yz+xz\le x^2+y^2+z^2\)
vì \(x^2+y^2\ge2xy\)
\(y^2+z^2\ge2yz\)
\(x^2+z^2\ge2xz\)
\(=>2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)< =>xy+yz+xz\le x^2+y^2+z^2\)
\(=>2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)\le10\left(x^2+y^2+z^2\right)\)
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{10\left(x^2+y^2+z^2\right)}=\dfrac{x^2+y^2+z^2}{10}=\dfrac{\dfrac{1}{3}}{10}=\dfrac{1}{30}\left(đpcm\right)\)
dấu"=" xảy ra<=>x=y=z=1/3
Áp dụng BĐT \(AM-GM\) ta có :
\(\left\{{}\begin{matrix}x^2+y^2\ge2xy\\y^2+3\ge2y+2\end{matrix}\right.\Rightarrow x^2+2y^2+3\ge2\left(xy+y+1\right)\Rightarrow\dfrac{1}{x^2+2y^2+3}\le\dfrac{1}{2\left(xy+y+1\right)}\)
Tương tự : \(\dfrac{1}{y^2+2z^2+3}\le\dfrac{1}{2\left(yz+z+1\right)}\)
\(\dfrac{1}{z^2+2x^2+3}\le\dfrac{1}{2\left(zx+x+1\right)}\)
Cộng từng vế BĐT ta được :
\(\dfrac{1}{x^2+2y^2+3}+\dfrac{1}{y^2+2z^2+3}+\dfrac{1}{z^2+2x^2+3}\le\dfrac{1}{2}\left(\dfrac{1}{xy+y+1}+\dfrac{1}{yz+z+1}+\dfrac{1}{zx+x+1}\right)=\dfrac{1}{2}\left(\dfrac{xyz}{xy+y+xyz}+\dfrac{x}{xyz+zx+x}+\dfrac{1}{zx+x+1}\right)=\dfrac{1}{2}\left(\dfrac{xz+x+1}{xy+x+1}\right)=\dfrac{1}{2}.1=\dfrac{1}{2}\)