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28 tháng 7 2018

\(\dfrac{a^3}{b^3}+\dfrac{a^3}{b^3}+1+\dfrac{b^3}{c^3}+\dfrac{b^3}{c^3}+1+\dfrac{c^3}{a^3}+\dfrac{c^3}{a^3}+1\ge3\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\)

\(\Leftrightarrow2\left(\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\right)\ge3\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)-3\)

\(\ge2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)+3-3=2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\)

\(\Leftrightarrow\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\)

27 tháng 7 2018

giả sử \(a>b>c>0\) thì ta có :

\(\dfrac{a^2}{b^2}\left(\dfrac{a}{b}-1\right)+\dfrac{b^2}{c^2}\left(\dfrac{b}{c}-1\right)+\dfrac{c^2}{a^2}\left(\dfrac{c}{a}-1\right)\ge2\dfrac{a}{b}+\dfrac{c^2}{a^2}\left(\dfrac{c}{a}-1\right)\)

\(=\dfrac{2a}{b}+\dfrac{c^3}{a^3}-\dfrac{c^2}{a^2}\ge0\)

làm tương tự cho trường hợp \(c>b>a>0\) ; \(b>a>c\)\(b>c>a\)

\(\Rightarrow\left(đpcm\right)\)

29 tháng 7 2018

\(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2\ge5\sqrt[5]{\dfrac{a^{20}b^2}{b^{12}}}=5.\dfrac{a^4}{b^2}\)

\(\Rightarrow4.\dfrac{a^5}{b^3}+b^2\ge5.\dfrac{a^4}{b^2}\)

Tương tự: \(4.\dfrac{b^5}{c^3}+c^2\ge5\dfrac{b^4}{c^2};4\dfrac{c^5}{a^3}+a^2\ge5.\dfrac{c^4}{a^2}\)

\(\Rightarrow4\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+a^2+b^2+c^2\ge5\left(\dfrac{c^4}{a^2}+\dfrac{a^4}{b^2}+\dfrac{b^4}{c^2}\right)\)

Lại có: \(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2+b^2+b^2\ge5a^2\)

\(\Rightarrow2.\dfrac{a^5}{b^3}+3b^2\ge5a^2\), tương tự: \(2.\dfrac{b^5}{c^3}+3c^2\ge5b^2;2\dfrac{c^5}{a^3}+3a^2\ge5c^2\)

\(\Rightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\ge a^2+b^2+c^2\)

\(\Rightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}+4.\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge4.\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+a^2+b^2+c^2\ge5.\left(\dfrac{c^4}{a^2}+\dfrac{a^4}{b^2}+\dfrac{b^4}{c^2}\right)\)

\(\Rightarrow dpcm\)

25 tháng 7 2018

giả sử \(a>b>c>0\) thì ta có :

\(\dfrac{a^4}{b^2}\left(\dfrac{a}{b}-1\right)+\dfrac{b^4}{c^2}\left(\dfrac{b}{c}-1\right)+\dfrac{c^4}{a^2}\left(\dfrac{c}{a}-1\right)\ge\dfrac{2a^2b}{c}+\dfrac{c^5}{a^3}-\dfrac{c^4}{a^2}\)

\(\ge\dfrac{2c^4b}{a}-\dfrac{c^4}{a^2}=\dfrac{c^4}{a}\left(2b-\dfrac{1}{a}\right)>0\)

làm tương tự cho trường hợp \(c>b>a>0\) ; \(b>a>c\)\(b>c>a\)

\(\Rightarrow\left(đpcm\right)\)

mấy câu cậu câu đăng khác bn làm tương tự nha . nếu bn lm không được thì có j mk lm luôn cho còn h mk bạn rồi :(

17 tháng 2 2018

áp dụng bdt côsi \(\dfrac{a^2}{b^3}+\dfrac{1}{a}+\dfrac{1}{a}\ge\dfrac{3}{b}\)

tuông tu \(\dfrac{b^2}{c^3}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{3}{c}\)

\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{3}{a}\)

suy ra vt +\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

suy ra dpcm

dau = xay ra khi a=b=c

7 tháng 5 2017

Theo hệ quả của bất đẳng thức Cauchy

\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)

\(\Leftrightarrow\dfrac{a^2}{2}+\dfrac{b^2}{c}+\dfrac{c^2}{c}\ge\dfrac{ab}{2}+\dfrac{bc}{2}+\dfrac{ca}{2}\)

\(\Leftrightarrow a^2-\dfrac{a^2}{2}+b^2-\dfrac{b^2}{2}+c^2-\dfrac{c^2}{2}\ge\dfrac{ab}{2}+\dfrac{bc}{2}+\dfrac{ca}{2}\)

\(\Leftrightarrow a^2+b^2+c^2\ge\dfrac{a^2+b^2+c^2+ab+bc+ca}{2}\)

\(\Leftrightarrow a^2+b^2+c^2\ge\dfrac{2\left(a^2+b^2+c^2+ab+bc+ca\right)}{4}\)

\(\Leftrightarrow a^2+b^2+c^2\ge\dfrac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}{4}\) (1)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow a^2+b^2\ge2\sqrt{a^2b^2}=2ab\)

\(\Rightarrow\left(a+b\right)^2\ge4ab\)

Tương tự ta có \(\left\{{}\begin{matrix}\left(b+c\right)^2\ge4bc\\\left(c+a\right)^2\ge4ca\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)^2c+\left(a+b\right)^2\ge4abc+\left(a+b\right)^2\\\left(b+c\right)^2a+\left(b+c\right)^2\ge4abc+\left(b+c\right)^2\\\left(c+a\right)^2b+\left(c+a\right)^2\ge4abc+\left(c+a\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^2\left(c+1\right)\ge4abc+\left(a+b\right)^2\\\left(b+c\right)^2\left(a+1\right)\ge4abc+\left(b+c\right)^2\\\left(c+a\right)^2\left(b+1\right)\ge4abc+\left(c+a\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{\left(a+b\right)^2\left(c+1\right)}\le\dfrac{8}{4abc+\left(a+b\right)^2}\\\dfrac{8}{\left(b+c\right)^2\left(a+1\right)}\le\dfrac{8}{4abc+\left(b+c\right)^2}\\\dfrac{8}{\left(c+a\right)^2\left(b+1\right)}\le\dfrac{8}{4abc+\left(c+a\right)^2}\end{matrix}\right.\) (2)

Từ (1) và (2)

\(\Rightarrow VT\ge\dfrac{8}{\left(a+b\right)^2\left(c+1\right)}+\dfrac{8}{\left(b+c\right)^2\left(a+1\right)}+\dfrac{8}{\left(c+a\right)^2\left(b+1\right)}+\dfrac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a^2\right)}{4}\) (3)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\dfrac{8}{\left(a+b\right)^2\left(c+1\right)}+\dfrac{\left(a+b\right)^2}{4}\ge2\sqrt{\dfrac{2}{c+1}}=\dfrac{4}{\sqrt{2\left(c+1\right)}}\)

Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{8}{\left(b+c\right)^2\left(a+1\right)}+\dfrac{\left(b+c\right)^2}{4}\ge\dfrac{4}{\sqrt{2\left(a+1\right)}}\\\dfrac{8}{\left(c+a\right)^2\left(b+1\right)}+\dfrac{\left(c+a\right)^2}{4}\ge\dfrac{4}{\sqrt{2\left(b+1\right)}}\end{matrix}\right.\)

\(\Rightarrow\dfrac{8}{\left(a+b\right)^2\left(c+1\right)}+\dfrac{8}{\left(b+c\right)^2\left(a+1\right)}+\dfrac{8}{\left(c+a\right)^2\left(b+1\right)}+\dfrac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a^2\right)}{4}\ge\dfrac{4}{\sqrt{2\left(c+1\right)}}+\dfrac{4}{\sqrt{2\left(a+1\right)}}+\dfrac{4}{\sqrt{2\left(b+1\right)}}\)(4)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\sqrt{2\left(c+1\right)}\le\dfrac{c+3}{2}\)

\(\Rightarrow\dfrac{4}{\sqrt{2\left(c+1\right)}}\ge\dfrac{8}{c+3}\)

Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2\left(a+1\right)}}\ge\dfrac{8}{a+3}\\\dfrac{4}{\sqrt{2\left(b+1\right)}}\ge\dfrac{8}{b+3}\end{matrix}\right.\)

\(\Rightarrow\dfrac{4}{\sqrt{2\left(c+1\right)}}+\dfrac{4}{\sqrt{2\left(a+1\right)}}+\dfrac{4}{\sqrt{2\left(b+1\right)}}\ge\dfrac{8}{a+3}+\dfrac{8}{b+3}+\dfrac{8}{c+3}\) (5)

Từ điều (3) , (4) , (5)

\(\Rightarrow\dfrac{8}{\left(a+b\right)^2+4abc}+\dfrac{8}{\left(b+c\right)^2+4abc}+\dfrac{8}{\left(c+a\right)^2+4abc}+a^2+b^2+c^2\ge\dfrac{8}{a+3}+\dfrac{8}{b+3}+\dfrac{8}{c+3}\) ( đpcm )

25 tháng 4 2018

Tớ chưa học bđt Cauchy-Schwwarz và hệ quả AM-GM thì sao?

10 tháng 5 2018

\(\dfrac{a^3}{b+c}+\dfrac{b^3}{a+c}+\dfrac{c^3}{a+b}\)

\(=\dfrac{a^4}{ab+ac}+\dfrac{b^4}{ab+bc}+\dfrac{c^4}{ac+bc}\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ac\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)}\)

\(=\dfrac{a^2+b^2+c^2}{2}=\dfrac{1}{2}\)

Dấu "=" xảy ra khi: \(a=b=c=\dfrac{1}{\sqrt{3}}\)

NV
21 tháng 2 2019

\(A=\dfrac{a^4}{a\left(b+c\right)}+\dfrac{b^4}{b\left(a+c\right)}+\dfrac{c^4}{c\left(a+b\right)}\)

\(\Rightarrow A\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2ab+2ac+2bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+a^2+c^2+b^2+c^2}=\dfrac{a^2+b^2+c^2}{2}=1\)

\(\Rightarrow A_{min}=1\) khi \(\left[{}\begin{matrix}a=b=c=\dfrac{\sqrt{6}}{3}\\a=b=c=\dfrac{-\sqrt{6}}{3}\end{matrix}\right.\)

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