Ta có đánh giá: \(\frac{1}{x^2+x}\ge\frac{5-3x}{4}\) \(\forall x>0\)

Thật vậy, BĐT tương đương:

\(\Leftrightarrow4\ge\left(x^2+x\right)\left(5-3x\right)\)

\(\Leftrightarrow3x^3-2x^2-5x+4\ge0\)

\(\Leftrightarrow\left(x-1\right)^2\left(3x+4\right)\ge0\) (luôn đúng \(\forall x>0\))

Tương tự ta có: \(\frac{1}{y^2+y}\ge\frac{5-3y}{4}\) ; \(\frac{1}{z^2+z}\ge\frac{5-3z}{4}\)

Cộng vế với vế: \(P\ge\frac{15-3\left(x+y+z\right)}{4}=\frac{15-9}{4}=\frac{3}{2}\)

\(P_{min}=\frac{3}{2}\) khi \(x=y=z=1\)

Thiếu chứng minh điều kiện bằng j bạn ơi

ban ghi ro de bai duoc ko ? mik ko hieu de bai

? CM cái gì

Câu hỏi của Winkies:bạn tham khảo tại đây nhé!

Trả lời

Từ giả thiết x+y+z=xyz <=> 1/xy + 1/yz + 1/zx = 1

Khi đó: x/1+x² = \(\frac{1}{\frac{x}{\left(\frac{1}{z}+\frac{1}{y}\right)\left(\frac{1}{x}+\frac{1}{z}\right)}}\)\(=\frac{xyz}{\left(x+y\right)\left(x+z\right)}\)

Tương tự cho 2 cái còn lại ta có:\(\frac{y}{1+y^2}=\frac{xyz}{\left(y+x\right)\left(y+z\right)}\)

\(\frac{z}{1+z^2}=\frac{xyz}{\left(z+x\right)\left(z+y\right)}\)

Suy ra VT=\(\frac{xyz\left(y+z\right)+2xyz\left(z+x\right)+3xyz\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)\(=\frac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

ĐPCM

 Ta có:\(\frac{x}{1+x^2}=\frac{xyz}{yz+x^2yz}=\frac{xyz}{yz+x\left(xyz\right)}=\frac{xyz}{yz+x\left(x+y+z\right)}=\frac{xyz}{yz+x^2+xy+xz}=\frac{xyz}{y\left(x+z\right)+x\left(x+z\right)}\)

\(=\frac{xyz}{\left(x+z\right)\left(y+x\right)}\)

Chứng minh tương tự : \(\frac{2y}{1+y^2}=\frac{2xyz}{\left(y+z\right)\left(y+x\right)}\)

                                        \(\frac{3z}{1+z^2}=\frac{3xyz}{\left(x+z\right)\left(x+y\right)}\)

Khi đó VT \(=\frac{xyz}{\left(x+z\right)\left(y+x\right)}+\frac{2xyz}{\left(y+z\right)\left(y+x\right)}+\frac{3xyz}{\left(x+z\right)\left(z+y\right)}\)

\(=\frac{xyz\left[y+z+2\left(z+x\right)+3\left(x+y\right)\right]}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

\(=\frac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\left(đpcm\right)\)

( mình đang vội nên làm hơi tắt mong bạn thông cảm )

chep bai vui khong may ban :)) 

đầu tiên ta chứng minh với x,y,z,t bất kì thì:

\(\sqrt{x^2+y^2}+\sqrt{z^2+t^2}\ge\sqrt{\left(x+z\right)^2+\left(y+t\right)^2}\) (*)

thật vậy bđt (*) tương đương với: 

\(x^2+y^2+z^2+t^2+2\sqrt{\left(x^2+y^2\right)\left(z^2+t^2\right)}\ge x^2+2xz+z^2+y^2+2yt+t^2\)

\(\Leftrightarrow\sqrt{\left(x^2+y^2\right)\left(z^2+t^2\right)}\ge xz+yt\)

bđt trên đúng vì theo bđt bunhia cốp xki

\(\sqrt{\left(x^2+y^2\right)\left(z^2+t^2\right)}\ge\sqrt{\left(xz+yt\right)^2}=|xz+yt|\ge xz+yt\)

Áp dụng (*) ta có:

\(P=\sqrt{4+x^4}+\sqrt{4+y^4}+\sqrt{4+z^4}\ge\sqrt{\left(2+2\right)^2+\left(x^2+y^2\right)^2}+\sqrt{4+z^2}\)

\(\ge\sqrt{\left(2+2+2\right)^2+\left(x^2+y^2+z^2\right)^2}=\sqrt{36+\left(x^2+y^2+z^2\right)^2}\)

Ta có:

\(\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2+\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Rightarrow3x^2+3y^2+3z^2+3\ge2x+2y+2z+2xy+2yz+2zx=2.6=12\)

\(\Rightarrow x^2+y^2+z^2\ge3\Rightarrow P\ge\sqrt{36+3}=3\sqrt{5}\)

Dấu bằng xảy ra khi x=y=z=1

srweafgtseawref

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{y+z+1}{x}=\frac{x+z+2019}{y}=\frac{x+y-2020}{z}=\frac{y+z+1+x+z+2019+x+y-2020}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

\(\Rightarrow2=\frac{1}{x+y+z}\)\(\Rightarrow x+y+z=\frac{1}{2}\)

Ta có: 

​+) \(\frac{y+z+1}{x}=2\)\(\Rightarrow y+z+1=2x\)\(\Rightarrow x+y+z+1=3x\)\(\Rightarrow\frac{1}{2}+1=3x\)\(\Rightarrow3x=\frac{3}{2}\)\(\Rightarrow x=\frac{1}{2}\)

+) \(\frac{x+z+2019}{y}=2\)\(\Rightarrow x+z+2019=2y\)\(\Rightarrow x+y+z+2019=3y\)\(\Rightarrow\frac{1}{2}+2019=3y\)\(\Rightarrow3y=\frac{4039}{2}\)\(\Rightarrow y=\frac{4039}{6}\)

+) \(\frac{x+y-2020}{z}=2\)\(\Rightarrow x+y-2020=2z\)\(\Rightarrow x+y+z-2020=3z\)\(\Rightarrow\frac{1}{2}-2020=3z\)\(\Rightarrow3z=\frac{-4039}{2}\)\(\Rightarrow z=\frac{-4039}{6}\)

Lại có: \(A=2016x+y^{2017}+z^{2017}=2016.\frac{1}{2}+\left(\frac{4039}{6}\right)^{2017}+\left(\frac{-4039}{6}\right)^{2017}=4032+\left(\frac{4039}{6}\right)^{2017}-\left(\frac{4039}{6}\right)^{2017}=4032\)

\(Gt\Rightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\)

Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\Rightarrow ab+bc+ca=1\)

\(VT=\frac{2}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}+\frac{1}{\sqrt{1+z^2}}\)

\(=\frac{\frac{2}{x}}{\sqrt{\frac{1}{x^2}+1}}+\frac{\frac{1}{y}}{\sqrt{\frac{1}{y^2}+1}}+\frac{\frac{1}{z}}{\sqrt{\frac{1}{z^2}+1}}\)

\(=\frac{2a}{\sqrt{a^2+ab+bc+ca}}+\frac{b}{\sqrt{b^2+ab+bc+ca}}+\frac{c}{\sqrt{c^2+ab+bc+ca}}\)

\(=\sqrt{\frac{2a}{\left(a+b\right)}\cdot\frac{2a}{\left(a+c\right)}}+\sqrt{\frac{2b}{\left(b+a\right)}\cdot\frac{b}{2\left(b+c\right)}}\)\(+\sqrt{\frac{2c}{\left(c+a\right)}\cdot\frac{c}{2\left(c+b\right)}}\)

\(\le\frac{\frac{2a}{a+b}+\frac{2a}{a+c}+\frac{2b}{a+b}+\frac{b}{2\left(b+c\right)}+\frac{2c}{c+a}+\frac{c}{2\left(c+b\right)}}{2}=\frac{9}{4}\)