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\(\dfrac{a^3}{b^3}+\dfrac{a^3}{b^3}+1+\dfrac{b^3}{c^3}+\dfrac{b^3}{c^3}+1+\dfrac{c^3}{a^3}+\dfrac{c^3}{a^3}+1\ge3\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\)
\(\Leftrightarrow2\left(\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\right)\ge3\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)-3\)
\(\ge2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)+3-3=2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\)
\(\Leftrightarrow\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\)
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 \(c>b>a>0\) ; \(b>a>c\) và \(b>c>a\)
\(\Rightarrow\left(đpcm\right)\)
Áp dụng bất đẳng thức AM-GM cho 2 số dương ta có: \(\dfrac{a^2}{b}+b\ge2\sqrt{\dfrac{a^2b}{b}}=2\sqrt{a^2}=2a\)
Tương tự với các vế ta được: \(\left\{{}\begin{matrix}\dfrac{b^2}{c}+c\ge2b\\\dfrac{c^2}{a}+a\ge2c\end{matrix}\right.\)
Cộng theo vế: \(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}+a+b+c\ge2a+2b+2c\)
\(\Rightarrow\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+b+c\)
\(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2\ge5\sqrt[5]{\dfrac{a^{20}b^2}{b^{12}}}=5.\dfrac{a^4}{b^2}\)
\(\Rightarrow4.\dfrac{a^5}{b^3}+b^2\ge5.\dfrac{a^4}{b^2}\)
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: \(2.\dfrac{b^5}{c^3}+3c^2\ge5b^2;2\dfrac{c^5}{a^3}+3a^2\ge5c^2\)
\(\Rightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\ge a^2+b^2+c^2\)
\(\Rightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}+4.\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge4.\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+a^2+b^2+c^2\ge5.\left(\dfrac{c^4}{a^2}+\dfrac{a^4}{b^2}+\dfrac{b^4}{c^2}\right)\)
\(\Rightarrow dpcm\)
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 \(c>b>a>0\) ; \(b>a>c\) và \(b>c>a\)
\(\Rightarrow\left(đpcm\right)\)
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 :(
Lời giải:
Phải thêm điều kiện $a,b,c>0$ nữa bạn nhé
Ta có:
\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq a+b+c\)
\(\Leftrightarrow \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-a-b-c\geq 0\)
\(\Leftrightarrow \frac{a^2}{b}-(2a-b)+\frac{b^2}{c}-(2b-c)+\frac{c^2}{a}-(2c-a)\geq 0\)
\(\Leftrightarrow \frac{a^2-2ab+b^2}{b}+\frac{b^2-2bc+c^2}{c}+\frac{c^2-2ac+a^2}{a}\geq 0\)
\(\Leftrightarrow \frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a}\geq 0\)
(luôn đúng với mọi $a,b,c>0$)
Do đó ta có đpcm
Dấu "=" xảy ra khi $a=b=c$
Hoặc có thể sử dụng BĐT Cauchy như sau:
\(\frac{a^2}{b}+b\geq 2\sqrt{\frac{a^2}{b}.b}=2a\)
\(\frac{b^2}{c}+c\ge 2\sqrt{\frac{b^2}{c}.c}=2b\)
\(\frac{c^2}{a}+a\geq 2\sqrt{\frac{c^2}{a}.a}=2c\)
Cộng theo vế:
\(\Rightarrow \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+(a+b+c)\geq 2(a+b+c)\)
\(\Rightarrow \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq a+b+c\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Lời giải:
Ta có:
Nhân cả hai vế với $a+b+c$ , BĐT cần chứng minh tương đương với:
\(\frac{(a^2+b^2)(a+b+c)}{a+b}+\frac{(b^2+c^2)(a+b+c)}{b+c}+\frac{(c^2+a^2)(a+b+c)}{c+a}\leq 3(a^2+b^2+c^2)\)
\(\Leftrightarrow 2(a^2+b^2+c^2)+\frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq 3(a^2+b^2+c^2)\)
\(\Leftrightarrow \frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq a^2+b^2+c^2\)
\(\Leftrightarrow \frac{c(a+b)^2-2abc}{a+b}+\frac{a(b+c)^2-2abc}{b+c}+\frac{b(a+c)^2-2abc}{a+c}\leq a^2+b^2+c^2\)
\(\Leftrightarrow 2(ab+bc+ac)\leq a^2+b^2+c^2+2abc\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)\)
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Áp dụng BĐT Cauchy- Schwarz:
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\geq \frac{9}{2(a+b+c)}\)
\(\Rightarrow a^2+b^2+c^2+2abc\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\geq a^2+b^2+c^2+\frac{9abc}{a+b+c}\)
Ta cần chứng minh \(a^2+b^2+c^2+\frac{9abc}{a+b+c}\geq 2(ab+bc+ac)\)
\(\Leftrightarrow (a^2+b^2+c^2)(a+b+c)+9abc\geq 2(ab+bc+ac)(a+b+c)\)
\(\Leftrightarrow a^3+b^3+c^3+3abc\geq ab(a+b)+bc(b+c)+ca(a+c)\)
(luôn đúng theo BĐT Schur)
Do đó ta có đpcm.
Dấu bằng xảy ra khi $a=b=c$
làm sao để có 1 chuỗi các ý tưởng hoàn hảo vậy bn :)) mình nháp hoài rồi mà toàn mắc :v
\(A=\dfrac{a^4}{a\left(b+c\right)}+\dfrac{b^4}{b\left(a+c\right)}+\dfrac{c^4}{c\left(a+b\right)}\)
\(\Rightarrow A\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2ab+2ac+2bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+a^2+c^2+b^2+c^2}=\dfrac{a^2+b^2+c^2}{2}=1\)
\(\Rightarrow A_{min}=1\) khi \(\left[{}\begin{matrix}a=b=c=\dfrac{\sqrt{6}}{3}\\a=b=c=\dfrac{-\sqrt{6}}{3}\end{matrix}\right.\)
Bạn chép đề sai?
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 )