Bài 2 : đã cm bên kia

Bài 1: :| 

we had điều này:

\(2=\frac{2014}{x}+\frac{2014}{y}+\frac{2014}{z}\)

\(\Leftrightarrow\frac{x-2014}{x}+\frac{y-2014}{y}+\frac{z-204}{z}=1\)

Xòng! bunyakovsky

P/s : Bệnh lười kinh niên tái phát nên ít khi ol sorry :<

1.

Áp dụng BĐT \(x^2+y^2+z^2\ge xy+yz+zx\)

\(\Rightarrow\left(\sqrt{ab}\right)^2+\left(\sqrt{bc}\right)^2+\left(\sqrt{ca}\right)^2\ge\sqrt{ab}.\sqrt{bc}+\sqrt{ab}.\sqrt{ac}+\sqrt{bc}.\sqrt{ac}\)

\(\Rightarrow ab+bc+ca\ge\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)

2.

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt[]{\frac{ab.bc}{ca}}=2b\) ; \(\frac{ab}{c}+\frac{ac}{b}\ge2a\) ; \(\frac{bc}{a}+\frac{ac}{b}\ge2c\)

Cộng vế với vế:

\(2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\right)\ge2\left(a+b+c\right)\)

\(\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c\)

3.

Từ câu b, thay \(c=1\) ta được:

\(ab+\frac{b}{a}+\frac{a}{b}\ge a+b+1\)

4.

\(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ac}\ge\frac{\left(a^2+b^2+c^2\right)}{ab+bc+ca}\)

\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge\frac{\left(ab+bc+ca\right)^2}{ab+bc+ca}=ab+bc+ca\)

Dấu "=" xảy ra khi \(a=b=c\)

5.

\(\frac{a}{bc}+\frac{b}{ca}\ge2\sqrt{\frac{ab}{bc.ca}}=\frac{2}{c}\) ; \(\frac{a}{bc}+\frac{c}{ab}\ge\frac{2}{b}\) ; \(\frac{b}{ca}+\frac{c}{ab}\ge\frac{2}{a}\)

Cộng vế với vế:

\(2\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\Rightarrow\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

1. bđt được viết lại thành

\(ab+bc+ca\ge a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\)

Theo bđt AM-GM thì :

\(ab+bc\ge2\sqrt{ab\cdot bc}=2\sqrt{ab^2c}=2b\sqrt{ac}\)

Tương tự : \(bc+ca\ge2c\sqrt{ab}\); \(ab+ca\ge2a\sqrt{bc}\)

Cộng vế với vế

=> \(2\left(ab+bc+ca\right)\ge2\left(a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\right)\)

=> \(ab+bc+ca\ge a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\)( đpcm )

Dấu "=" xảy ra <=> a=b=c

xài mincopski thử, tui ăn cơm đã

#: Lỡ hẹn với Mincopxki rồi xài cách khác vậy :(

Đặt \(a=\frac{2x}{3};b=\frac{2y}{3};c=\frac{2z}{3}\)

Khi đó ta có \(xy+yz+xz\ge3\) và cần chứng minh

\(Σ_{cyc}\sqrt{\frac{4x^2}{9}+\frac{9}{\left(2y+3\right)^2}}\ge\frac{\sqrt{181}}{5}\)

Áp dụng BĐT Cauchy-Schwarz ta có:\(Σ_{cyc}\sqrt{\frac{4x^2}{9}+\frac{9}{\left(2y+3\right)^2}}\)

\(=\frac{15}{\sqrt{181}}Σ_{cyc}\sqrt{\left(\frac{4}{9}+\frac{9}{25}\right)\left(\frac{4x^2}{9}+\frac{9}{\left(2y+3\right)^2}\right)}\ge\frac{15}{\sqrt{181}}Σ_{cyc}\left(\frac{4x}{9}+\frac{9}{5\left(2y+3\right)}\right)\)

Giờ chỉ cần chứng minh \(\frac{15}{\sqrt{181}}Σ_{cyc}\left(\frac{4x}{9}+\frac{9}{5\left(2y+3\right)}\right)\ge\frac{\sqrt{181}}{5}\)

\(\Leftrightarrow20\left(x+y+z\right)+81\left(\frac{1}{2x+3}+\frac{1}{2y+3}+\frac{1}{2z+3}\right)\ge\frac{543}{5}\)

Đặt tiếp \(x+y+z=3u;xy+yz+xz=3v^2\left(v>0\right)\)

Vì thế \(u\ge v\ge1\)và áp dụng BĐT C-S dạng Engel ta có:

\(20\left(x+y+z\right)+81\left(\frac{1}{2x+3}+\frac{1}{2y+3}+\frac{1}{2z+3}\right)-\frac{543}{5}\)

\(\ge20\left(x+y+z\right)+81\cdot\frac{\left(1+1+1\right)^2}{Σ_{cyc}\left(2x+3\right)}-\frac{543}{5}=60u+\frac{729}{6u+9}-\frac{543}{5}\)

\(=3\left(20u+\frac{81}{2u+3}-\frac{181}{5}\right)=\frac{6\left(u-1\right)\left(100u+69\right)}{5\left(2u+3\right)}\ge0\) 

Điều này đúng tức là ta có ĐPCM

a) \(\frac{a+b}{2}\ge\sqrt{ab}\)

\(\Leftrightarrow\frac{a^2+2ab+b^2}{4}-ab\ge0\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng \(\forall a,b\) )

=>đpcm

Cô si

\(\frac{bc}{a}+\frac{ca}{b}\ge2\sqrt{\frac{bc}{a}\cdot\frac{ca}{b}}=2c\)

\(\frac{ca}{b}+\frac{ab}{c}\ge2\sqrt{\frac{ca}{b}\cdot\frac{ab}{c}}=2a\)

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}\cdot\frac{bc}{a}}=2b\)

Cộng lại ta có:

\(2\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)\ge2\left(a+b+c\right)\Rightarrowđpcm\)

quên làm câu c:(

\(12=3a+5b\ge2\sqrt{15ab}\Rightarrow\sqrt{15ab}\le6\Rightarrow15ab\le36\Rightarrow ab\le\frac{36}{15}\)

Dấu "=" xảy ra tại \(3a=5b\Leftrightarrow\frac{a}{5}=\frac{b}{3}\)

1) \(\Sigma\frac{a}{b^3+ab}=\Sigma\left(\frac{1}{b}-\frac{b}{a+b^2}\right)\ge\Sigma\frac{1}{a}-\Sigma\frac{1}{2\sqrt{a}}=\Sigma\left(\frac{1}{a}-\frac{2}{\sqrt{a}}+1\right)+\Sigma\frac{3}{2\sqrt{a}}-3\)

\(\ge\Sigma\left(\frac{1}{\sqrt{a}}-1\right)^2+\frac{27}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}-3\ge\frac{27}{2\sqrt{3\left(a+b+c\right)}}-3=\frac{3}{2}\)

2.

Vỉ \(ab+bc+ca+abc=4\)thi luon ton tai \(a=\frac{2x}{y+z};b=\frac{2y}{z+x};c=\frac{2z}{x+y}\)

\(\Rightarrow VT=2\Sigma_{cyc}\sqrt{\frac{ab}{\left(b+c\right)\left(c+a\right)}}\le2\Sigma_{cyc}\frac{\frac{b}{b+c}+\frac{a}{c+a}}{2}=3\)

Cho o dong 2 la x,y,z nhe,ghi nham

\(\frac{a+b}{2}\ge\sqrt{ab}\)

\(\Leftrightarrow\frac{a+b}{2}-\sqrt{ab}\ge0\)

\(\Leftrightarrow\frac{a+b-2\sqrt{ab}}{2}\ge0\)

\(\Leftrightarrow\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2}\ge0\) (luôn đúng)

Vậy \(\frac{a+b}{2}\ge\sqrt{ab}\) (1)

\(\sqrt{ab}\ge\frac{2}{\frac{1}{a}+\frac{1}{b}}\)

\(\Leftrightarrow\sqrt{ab}\ge\frac{2ab}{a+b}\)

\(\Leftrightarrow\sqrt{ab}\ge\frac{2\sqrt{ab}^2}{a+b}\)

\(\Leftrightarrow\frac{2\sqrt{ab}}{a+b}\le1\)

\(\Leftrightarrow\frac{2\sqrt{ab}}{a+b}-1\le0\)

\(\Leftrightarrow\frac{2\sqrt{ab}-a-b}{a+b}\le0\)

\(\Leftrightarrow\frac{-\left(\sqrt{a}-\sqrt{b}\right)^2}{a+b}\le0\) (luôn đúng)

Vậy \(\sqrt{ab}\ge\frac{2}{\frac{1}{a}+\frac{1}{b}}\) (2)

Từ (1) ; (2) \(\Rightarrow\frac{a+b}{2}\ge\sqrt{ab}\ge\frac{2}{\frac{1}{a}+\frac{1}{b}}\) (đpcm)

\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{ab+1}\)

\(\Leftrightarrow\frac{1}{a^2+1}-\frac{1}{ab+1}+\frac{1}{b^2+1}-\frac{1}{ab+1}\ge0\)

\(\Leftrightarrow\frac{ab+1-a^2-1}{\left(a^2+1\right)\left(ab+1\right)}+\frac{ab+1-b^2-1}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)

\(\Leftrightarrow\frac{\left(ab-a^2\right)\left(b^2+1\right)+\left(ab-b^2\right)\left(a^2+1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)

\(\Leftrightarrow\frac{-a\left(b^2+1\right)\left(a-b\right)+b\left(a-b\right)\left(a^2+1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)\left(-ab^2-a+a^2b+b\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)\left[ab\left(a-b\right)-\left(a-b\right)\right]}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2\left(ab-1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)( luôn đúng )

Dấu "=" xảy ra \(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\ab-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=b\\ab=1\end{matrix}\right.\)