Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\sum\dfrac{x}{x^2+1}\le\sum\dfrac{x}{2x}\le\dfrac{3}{2}\)
GTLN là \(\dfrac{3}{2}\Leftrightarrow x=y=z=1\)
Áp dung BĐT AM-GM ta có
\(P=\dfrac{x^2}{x^4+yz}+\dfrac{y^2}{y^4+xz}+\dfrac{z^2}{z^4+xy}\)
\(\le\dfrac{x^2}{2x^2\sqrt{yz}}+\dfrac{y^2}{2y^2\sqrt{xz}}+\dfrac{z^2}{2z^2\sqrt{xy}}\)
\(=\dfrac{1}{2\sqrt{yz}}+\dfrac{1}{2\sqrt{xz}}+\dfrac{1}{2\sqrt{xy}}\)
\(\le\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{1}{2}\cdot\dfrac{xy+yz+xz}{xyz}\)
\(\le\dfrac{1}{2}\cdot\dfrac{x^2+y^2+z^2}{xyz}\le\dfrac{1}{2}\cdot\dfrac{3xyz}{xyz}=\dfrac{3}{2}\)
Dấu "=" <=> \(x=y=z=1\)
Từ GT ta có: \(3=\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}\ge\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\)
Suy ra \(3\le x+y+z\)
Áp dụng AM-GM:
\(VT\le\dfrac{x^2}{2x^2\sqrt{yz}}+\dfrac{y^2}{2y^2\sqrt{xz}}+\dfrac{z^2}{2z^2\sqrt{xy}}=\dfrac{1}{2}\sum\dfrac{1}{\sqrt{xy}}\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}{2\sqrt{xyz}}\le\dfrac{\sqrt{3\left(x+y+z\right)}}{2\sqrt{xyz}}\le\dfrac{1}{2}\sqrt{\dfrac{\left(x+y+z\right)^2}{xyz}}\)
\(\le\dfrac{1}{2}\sqrt{\dfrac{3\left(x^2+y^2+z^2\right)}{xyz}}=\dfrac{3}{2}\)
Vậy \(P_{Max}=\dfrac{3}{2}\)
Ta co : (x+y)2≤2(x2+y2)
=> x+y≤\(\sqrt{2\left(x^2+y^2\right)}\)
=> \(\dfrac{z^2}{x+y}\ge\dfrac{z^2}{\sqrt{2\left(x^2+y^2\right)}}\)
Tuong tu: \(\dfrac{x^2}{y+z}\ge\dfrac{x^2}{\sqrt{2\left(y^2+z^2\right)}}\)
\(\dfrac{y^2}{x+z}\ge\dfrac{y^2}{\sqrt{2\left(x+z\right)}}\)
VT≥\(\dfrac{x^2}{\sqrt{2\left(y^2+z^2\right)}}+\dfrac{y^2}{\sqrt{2\left(x^2+z^2\right)}}+\dfrac{z^2}{\sqrt{2\left(x^2+y^2\right)}}\)
Dat : \(\sqrt{y^2+z^2}=a\)
\(\sqrt{x^2+z^2}=b\)
\(\sqrt{x^2+y^2}=c\)
=> a+b+c=2015 , a2=y2+z2 , b2=x2+z2 , c2=x2+y2
=> VT≥ \(\dfrac{b^2+c^2-a^2}{2\sqrt{2}.a}+\dfrac{a^2+c^2-b^2}{2\sqrt{2}.b}+\dfrac{a^2+b^2-c^2}{2\sqrt{2}c}\)
≥ \(\dfrac{1}{2\sqrt{2}}\left[\dfrac{\left(b+c\right)^2}{2a}+\dfrac{\left(a+b\right)^2}{2c}+\dfrac{\left(a+c\right)^2}{2b}-2015\right]\)
≥\(\dfrac{1}{2\sqrt{2}}\left[2\left(a+b+c\right)-2015\right]\)
= \(\dfrac{2015}{2\sqrt{2}}\)
Ta sẽ cm bđt:\(\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}\ge\dfrac{\left(x+y+z\right)^2}{a+b+c}\)
Áp dụng bđt bunhia:
\(\left(\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}\right)\left(a+b+c\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}\ge\dfrac{\left(x+y+z\right)^2}{a+b+c}\)
Áp dụng vào suy ra:
\(A=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{2}{2}=1\)
"="<=>x=y=z=\(\dfrac{2}{3}\)
Cách khác:
Áp dụng BĐT Cauchy cho các số dương ta có:
\(\frac{x^2}{y+z}+\frac{y+z}{2}\geq 2\sqrt{\frac{x^2}{y+z}.\frac{y+z}{4}}=x\)
\(\frac{y^2}{x+z}+\frac{x+z}{4}\geq 2\sqrt{\frac{y^2}{x+z}.\frac{x+z}{4}}=y\)
\(\frac{z^2}{x+y}+\frac{x+y}{4}\geq 2\sqrt{\frac{z^2}{x+y}.\frac{x+y}{4}}=z\)
Cộng theo vế và rút gọn ta có:
\(A+\frac{x+y+z}{2}\geq x+y+z\)
\(\Rightarrow A\geq \frac{x+y+z}{2}=1\)
Vậy \(A_{\min}=1\Leftrightarrow x=y=z=\frac{2}{3}\)
Ta có: \(\dfrac{x}{x^2+1+y^2+1}\le\dfrac{x}{2\sqrt{\left(x^2+1\right)\left(y^2+1\right)}}\le\dfrac{1}{4}\left(\dfrac{x^2}{x^2+1}+\dfrac{1}{y^2+1}\right)\)
Tương tự: \(\dfrac{y}{y^2+z^2+2}\le\dfrac{1}{4}\left(\dfrac{y^2}{y^2+1}+\dfrac{1}{z^2+1}\right)\) ; \(\dfrac{z}{z^2+x^2+2}\le\dfrac{1}{4}\left(\dfrac{z^2}{z^2+1}+\dfrac{1}{x^2+1}\right)\)
Cộng vế với vế:
\(P\le\dfrac{1}{4}\left(\dfrac{x^2}{x^2+1}+\dfrac{1}{x^2+1}+\dfrac{y^2}{y^2+1}+\dfrac{1}{y^2+1}+\dfrac{z^2}{z^2+1}+\dfrac{1}{z^2+1}\right)=\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(x=y=z=1\)