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\(VT=\left(x^2+y^2+z^2\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)=3+\dfrac{x^2+y^2}{z^2}+z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\)
\(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}>=2\cdot\sqrt{\dfrac{y^2}{x^2}\cdot\dfrac{x^2}{y^2}}=2\)
=>\(VT>=5+\left(\dfrac{x^2}{z^2}+\dfrac{z^2}{16x^2}\right)+\left(\dfrac{y^2}{z^2}+\dfrac{z^2}{16y^2}\right)+\dfrac{15}{16}z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\)
\(\dfrac{x^2}{z^2}+\dfrac{z^2}{16x^2}>=2\cdot\sqrt{\dfrac{x^2}{z^2}\cdot\dfrac{z^2}{16x^2}}=\dfrac{1}{2}\)
\(\dfrac{y^2}{z^2}+\dfrac{z^2}{16y^2}>=\dfrac{1}{2}\)
và \(\dfrac{1}{x^2}+\dfrac{1}{y^2}>=\dfrac{2}{xy}>=\dfrac{2}{\left(\dfrac{x+y}{2}\right)^2}=\dfrac{8}{\left(x+y\right)^2}\)
=>\(\dfrac{15}{16}z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)>=\dfrac{15}{16}z^2\cdot\dfrac{8}{\left(x+y\right)^2}=\dfrac{15}{2}\left(\dfrac{z}{x+y}\right)^2=\dfrac{15}{2}\)
=>VT>=5+1/2+1/2+15/2=27/2
\(\sum\dfrac{x^4y}{x^2+1}=\sum\dfrac{x^3.\dfrac{1}{z}}{x^2+xyz}=\sum\dfrac{x^2}{z\left(x+yz\right)}=\sum\dfrac{x^2}{xz+1}\)
Áp dụng bất đẳng thức cauchy-schwarz:
\(Vt=\sum\dfrac{x^2}{xz+1}\ge\dfrac{\left(x+y+z\right)^2}{xy+yz+xz+3}\)
mà theo AM-GM: \(xy+yz+xz\ge3\sqrt[3]{x^2y^2z^2}=3\)
hay \(3\le xy+yz+xz\)
do đó \(VT\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\ge\dfrac{3\left(xy+yz+zx\right)}{2\left(xy+yz+xz\right)}=\dfrac{3}{2}\)
Dấu = xảy ra khi x=y=z=1
P/s: Câu này khoai
Giả thiết thiếu rồi em, chỗ \(\dfrac{1}{x+1}+...\) thiếu đoạn sau nữa
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)
\(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}=\dfrac{x^2+y^2+2}{\left(xy\right)^2+x^2+y^2+1}=1-\dfrac{\left(xy\right)^2-1}{\left(xy\right)^2+x^2+y^2+1}\ge1-\dfrac{\left(xy\right)^2-1}{\left(xy\right)^2+2xy+1}\)
\(\Rightarrow\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}\ge1-\dfrac{\left(xy+1\right)\left(xy-1\right)}{\left(xy+1\right)^2}=1-\dfrac{xy-1}{xy+1}=\dfrac{2}{1+xy}\) (đpcm)
b. Tương tự câu a:
\(\dfrac{1}{1+x^2}+\dfrac{1}{1+z^2}\ge\dfrac{2}{1+zx}\) ; \(\dfrac{1}{1+y^2}+\dfrac{1}{1+z^2}\ge\dfrac{2}{1+yz}\)
Cộng vế với vế và rút gọn:
\(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}+\dfrac{1}{1+z^2}\ge\dfrac{1}{1+xy}+\dfrac{1}{1+yz}+\dfrac{1}{z+zx}\) (1)
Mà \(\left\{{}\begin{matrix}z\ge1\Rightarrow1+xy\le1+xyz\\y\ge1\Rightarrow1+zx\le1+xyz\\x\ge1\Rightarrow1+yz\le1+xyz\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{1+xy}+\dfrac{1}{1+yz}+\dfrac{1}{1+zx}\ge\dfrac{1}{1+xyz}+\dfrac{1}{1+xyz}+\dfrac{1}{1+xyz}=\dfrac{3}{1+xyz}\) (2)
TỪ (1); (2) \(\Rightarrowđpcm\)
a) Ta có: \(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}\ge\dfrac{2}{1+xy}\)
\(\Leftrightarrow\dfrac{1}{1+x^2}-\dfrac{1}{1+xy}+\dfrac{1}{1+y^2}-\dfrac{1}{1+xy}\ge0\)
\(\Leftrightarrow\dfrac{\left(1+xy\right)-\left(1+x^2\right)}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{\left(1+xy\right)-\left(1+y^2\right)}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(xy-x^2\right)\left(1+y^2\right)+\left(xy-y^2\right)\left(1+x^2\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow\dfrac{xy+xy^3-x^2-x^2y^2+xy+x^3y-y^2-x^2y^2}{\left(1+xy\right)\left(1+x^2\right)\left(1+y^2\right)}\ge0\)
\(\Leftrightarrow\dfrac{2xy+xy\left(x^2+y^2\right)-2x^2y^2-x^2-y^2}{\left(1+xy\right)\left(1+x^2\right)\left(1+y^2\right)}\ge0\)
\(\Leftrightarrow\dfrac{xy\left(x^2-2xy+y^2\right)-\left(x^2-2xy+y^2\right)}{\left(1+xy\right)\left(1+y^2\right)\left(1+x^2\right)}\ge0\)
\(\Leftrightarrow\dfrac{xy\left(x-y\right)^2-\left(x-y\right)^2}{\left(1+xy\right)\left(1+x^2\right)\left(1+y^2\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(x-y\right)^2\left(xy-1\right)}{\left(1+xy\right)\left(1+x^2\right)\left(1+y^2\right)}\ge0\)(luôn đúng)
=> Đẳng thức ban đầu được chứng minh.
P/s: Cái đoạn sau bạn bổ sung thêm vào là vì x và y lớn hơn bằng 1 nên xy-1 sẽ lớn hơn hoặc bằng 0 nhé, mình lười quá ngại chèn:vv.
Còn câu b bạn đợi mình nháp xíu.
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 $x=y=z=1$
Xài Bunhiacopxki thì bài này sẽ hơi dài:
Đặt vế trái là P
Ta có:
\(\left(\dfrac{1}{4}+4\right)\left(x^2+\dfrac{1}{x^2}\right)\ge\left(\dfrac{x}{2}+\dfrac{2}{x}\right)^2\)
\(\Leftrightarrow\dfrac{17}{4}\left(x^2+\dfrac{1}{x^2}\right)\ge\left(\dfrac{x}{2}+\dfrac{2}{x}\right)^2\)
\(\Rightarrow\sqrt{x^2+\dfrac{1}{x^2}}\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{x}{2}+\dfrac{2}{x}\right)\)
Tương tự:
\(\sqrt{y^2+\dfrac{1}{y^2}}\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{y}{2}+\dfrac{2}{y}\right)\) ; \(\sqrt{z^2+\dfrac{1}{z^2}}\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{z}{2}+\dfrac{2}{z}\right)\)
Cộng vế: \(P\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{x}{2}+\dfrac{y}{2}+\dfrac{z}{2}+\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}\right)\)
\(P\ge\dfrac{1}{\sqrt{17}}\left(x+y+z+4\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\right)\ge\dfrac{1}{\sqrt{17}}\left(x+y+z+\dfrac{36}{x+y+z}\right)\)
\(P\ge\dfrac{1}{\sqrt{17}}\left(x+y+z+\dfrac{9}{4\left(x+y+z\right)}+\dfrac{135}{4\left(x+y+z\right)}\right)\)
\(P\ge\dfrac{1}{\sqrt{17}}\left(2\sqrt{\dfrac{9\left(x+y+z\right)}{4\left(x+y+z\right)}}+\dfrac{135}{4.\dfrac{3}{2}}\right)=\dfrac{3}{2}\sqrt{17}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{2}\)
Đề sai nhé, \(\dfrac{z^2}{x+1}\) mới đúng nha
\(\dfrac{x^2}{y+1}+\dfrac{y^2}{z+1}+\dfrac{z^2}{x+1}\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+3}\left(\text{Svácxơ}\right)\)
\(\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
Ta có: \(x+y+z\ge3\sqrt[3]{xyz}=3\)
\(\Rightarrow x+y+z+3\le2\left(x+y+z\right)\)