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\(\dfrac{x^3}{2y+1}+\dfrac{2y+1}{9}+\dfrac{1}{3}\ge3\sqrt[3]{\dfrac{x^3\left(2y+1\right)}{27\left(2y+1\right)}}=x\)
Tương tự: \(\dfrac{y^3}{2z+1}+\dfrac{2z+1}{9}+\dfrac{1}{3}\ge y\) ; \(\dfrac{z^3}{2x+1}+\dfrac{2x+1}{9}+\dfrac{1}{3}\ge z\)
Cộng vế:
\(VT+\dfrac{2\left(x+y+z\right)+3}{9}+1\ge x+y+z\)
\(\Rightarrow VT\ge\dfrac{7}{9}\left(x+y+z\right)-\dfrac{4}{3}\ge\dfrac{7}{9}.3\sqrt[3]{xyz}-\dfrac{4}{3}=1\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
Ta có \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xyz}=1\)
\(\Leftrightarrow\dfrac{\left(yz\right)^2+\left(xz\right)^2+\left(xy\right)^2+2xyz}{\left(xyz\right)^2}=1\)
<=> (xy)2 + (yz)2 + (zx)2 + 2xyz = (xyz)2
<=> (xy)2 + (yz)2 + (xz)2 + 2xyz(x + y + z) = (xyz)2
<=> (xy + yz + zx)2 = (xyz)2
<=> \(\left[{}\begin{matrix}xy+yz+zx=xyz\\xy+yz+zx=-xyz\end{matrix}\right.\)
+) Khi xy + yz + zx = -xyz
=> \(\dfrac{xy+yz+zx}{xyz}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=-1< 0\left(\text{loại}\right)\)
=> xy + yz + zx = xyz
<=> \(xyz\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=xyz\Leftrightarrow xyz\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-1\right)=0\)
<=> \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)
<=> \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)
<=> \(\dfrac{x+y}{xy}=\dfrac{-\left(x+y\right)}{\left(x+y+z\right)z}\)
<=> \(\left(x+y\right)\left(\dfrac{1}{xz+yz+z^2}+\dfrac{1}{xy}\right)=0\)
<=> \(\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{\left(zx+yz+z^2\right)xy}=0\)
<=> \(\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)
Khi x = -y => y = 1 => P = 1
Tương tự y = -z ; z = -x được P = 1
Vậy P = 1
Đặt vế trái của BĐT cần chứng minh là P
Ta có:
\(P=\dfrac{\sqrt{xy+\left(x+y+z\right)z}+\sqrt{2\left(x^2+y^2\right)}}{1+\sqrt{xy}}=\dfrac{\sqrt{\left(x+z\right)\left(y+z\right)}+\sqrt{2\left(x^2+y^2\right)}}{1+\sqrt{xy}}\)
\(P\ge\dfrac{\sqrt{\left(\sqrt{xy}+z\right)^2}+\sqrt{\left(x+y\right)^2}}{1+\sqrt{xy}}=\dfrac{\sqrt{xy}+x+y+z}{1+\sqrt{xy}}=\dfrac{\sqrt{xy}+1}{1+\sqrt{xy}}=1\) (đpcm)
Dấu "=" xảy ra khi \(x=y\)
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 $x=y=z=1$
Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :
\(\dfrac{1}{xy}+\dfrac{1}{xz}\ge\dfrac{\left(1+1\right)^2}{xy+xz}=\dfrac{4}{x\left(y+z\right)}\)(1)
Áp dụng bất đẳng thức AM-GM ta có :
\(x\left(y+z\right)\le\dfrac{\left(x+y+z\right)^2}{4}=4\)=> \(\dfrac{1}{x\left(y+z\right)}\ge\dfrac{1}{4}\)=> \(\dfrac{4}{x\left(y+z\right)}\ge1\)(2)
Từ (1) và (2) => \(\dfrac{1}{xy}+\dfrac{1}{xz}\ge\dfrac{4}{x\left(y+z\right)}\ge1\)=> \(\dfrac{1}{xy}+\dfrac{1}{xz}\ge1\)(đpcm)
Đẳng thức xảy ra <=> x = 2 ; y = z = 1
Giả thiết thiếu rồi em, chỗ \(\dfrac{1}{x+1}+...\) thiếu đoạn sau nữa
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 \(\left(\dfrac{1}{\sqrt{x}};\dfrac{1}{\sqrt{y}};\dfrac{1}{\sqrt{z}}\right)=\left(a;b;c\right)\Rightarrow\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}=1\)
Ta cần chứng minh: \(ab+bc+ca\le\dfrac{3}{2}\)
Thật vậy, ta có:
\(1=\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3}\)
\(\Rightarrow a^2+b^2+c^2+3\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Rightarrow ab+bc+ca\le\dfrac{3}{2}\) (đpcm)
Guể :v t nhớ làm bài này rồi mà :v
Đặt \(x=\dfrac{bc}{a^2};y=\dfrac{ac}{b^2};z=\dfrac{ab}{c^2}\)\(\Rightarrow\left\{{}\begin{matrix}abc=1\\a,b,c>0\end{matrix}\right.\)
Và \(BDT\Leftrightarrow\dfrac{a^4}{b^2c^2+a^2bc+a^4}+\dfrac{b^4}{a^2c^2+ab^2c+b^4}+\dfrac{c^4}{a^2b^2+abc^2+c^4}\ge1\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{b^2c^2+a^2bc+a^2c^2+ab^2c+a^2b^2+abc^2+a^4+b^4+c^4}\)
Cần chứng minh \(\dfrac{\left(a^2+b^2+c^2\right)^2}{b^2c^2+a^2bc+a^2c^2+ab^2c+a^2b^2+abc^2+a^4+b^4+c^4}\ge1\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2\ge b^2c^2+a^2bc+a^2c^2+ab^2c+a^2b^2+abc^2+a^4+b^4+c^4\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)\ge b^2c^2+a^2bc+a^2c^2+ab^2c+a^2b^2+abc^2+a^4+b^4+c^4\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2\ge ab^2c+a^2bc+abc^2\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2\ge abc\left(a+b+c\right)\) *Đúng theo AM-GM*
uh bài này làm rồi, tại lúc đó đầu hơi ngu nên không nhớ ra, thông cảm nhé