Cho a,b,c>0.Chứng minh rằng:
\(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\ge\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
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áp dụng dbt cosi cho 2 số:\(\frac{a^3}{b^2}\)va b ta duoc :
\(\frac{a^3}{b^2}\)+a\(\ge\)2\(\sqrt{\frac{a^3}{b^2}.a}\)=2\(\frac{a^2}{b}\)
CMTT:\(\frac{b^3}{c^2}\)+b\(\ge\)2\(\frac{b^2}{c}\)
\(\frac{c^3}{a^2}\)+c\(\ge\)2\(\frac{c^2}{a}\)
\(\Rightarrow\)\(\frac{a^3}{b^2}\)+\(\frac{b^3}{c^2}\)+\(\frac{c^3}{a^2}\)+(a+b+c)\(\ge\)2(\(\frac{a^2}{b}\)+\(\frac{b^2}{c}\)+\(\frac{c^2}{a}\))
\(\Leftrightarrow\)\(\frac{a^3}{b^2}\)+\(\frac{b^3}{c^2}\)+\(\frac{c^3}{a^2}\)\(\ge\)2(\(\frac{a^2}{b}\)+\(\frac{b^2}{c}\)+\(\frac{c^2}{a}\)) - (a+b+c) (1)
Ap dụng bdt cosi cho các số dương , ta được:
\(\frac{a^2}{b}\)+\(b\)\(\ge\)2\(\sqrt{\frac{a^2}{b}.b}\)=2a
CMTT: \(\frac{b^2}{c}\)+c\(\ge\)2b
\(\frac{c^2}{a}\)+a\(\ge\)2c
\(\Rightarrow\)\(\frac{a^2}{b}\)+\(\frac{b^2}{c}\)+\(\frac{c^2}{a}\)+(a+b+c) \(\ge\)2(a+b+c)
\(\Leftrightarrow\)\(\frac{a^2}{b}\)+\(\frac{b^2}{c}\)+\(\frac{c^2}{a}\)\(\ge\)a+b+c
\(\Leftrightarrow\)\(\frac{a^2}{b}\)+\(\frac{b^2}{c}\)+\(\frac{c^2}{a}\) _ (a + b + c ) \(\ge\)0
Do Đó:TỪ (1) ta co : \(\frac{a^3}{b^2}\)+\(\frac{b^3}{c^2}\)+\(\frac{b^3}{c^2}\)\(\ge\)(\(\frac{a^2}{b}\)+\(\frac{b^2}{c}\)+\(\frac{c^2}{a}\) )
Xét hiệu hai vế:
BĐT \(\Leftrightarrow\left(\frac{a^3}{b^2}-\frac{a^2b}{b^2}\right)+\left(\frac{b^3}{c^2}-\frac{b^2c}{c^2}\right)+\left(\frac{c^3}{a^2}-\frac{c^2a}{a^2}\right)-\left(a+b+c-b-c-a\right)\ge0\)
\(\Leftrightarrow\left(\frac{a^3}{b^2}-\frac{a^2b}{b^2}\right)+\left(\frac{b^3}{c^2}-\frac{b^2c}{c^2}\right)+\left(\frac{c^3}{a^2}-\frac{c^2a}{a^2}\right)-\left[\left(a-b\right)+\left(b-c\right)+\left(c-a\right)\right]\ge0\)
\(\Leftrightarrow\left(\frac{a^2}{b^2}\left(a-b\right)-\left(a-b\right)\right)+\left(\frac{b^2}{c^2}\left(b-c\right)-\left(b-c\right)\right)+\left(\frac{c^2}{a^2}\left(c-a\right)-\left(c-a\right)\right)\ge0\)
\(\Leftrightarrow\frac{\left(a+b\right)\left(a-b\right)^2}{b^2}+\frac{\left(b+c\right)\left(b-c\right)^2}{c^2}+\frac{\left(c+a\right)\left(c-a\right)^2}{a^2}\ge0\)
BĐT này đúng với mọi a,b,c > 0 nên ta có Q.E.D
Dấu "=" xảy ra khi a =b =c
P/s: Toán 7 gì mà khó thế nhỉ??Mình cũng không chắc đâu nha!
Áp dụng bất đẳng thức Cô-si ta có:
\(\dfrac{a^2}{b^3}+\dfrac{1}{a}+\dfrac{1}{a}\ge\sqrt[3]{\dfrac{a^2}{b^3}.\dfrac{1}{a}.\dfrac{1}{a}}=\dfrac{3}{b}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\sqrt[3]{\dfrac{c^2}{a^3}.\dfrac{1}{c}.\dfrac{1}{c}}=\dfrac{3}{a}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\sqrt[3]{\dfrac{c^2}{a^3}.\dfrac{1}{c}.\dfrac{1}{c}}=\dfrac{3}{a}\)
Cộng theo vế ta được:
\(\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{a^2}{a^3}+\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\ge3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{c^2}{a^3}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Áp dụng Svac
\(\Sigma\frac{a^3}{b+c}=\Sigma\frac{a^4}{ab+ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)}=\frac{1}{2}\left(a^2+b^2+c^2\right)\)
"=" tại a=b=c
E thử làm cách khác ạ:))
Không mất tính tổng quát,giả sử \(a\ge b\ge c\)
\(\Rightarrow\hept{\begin{cases}a^2\ge b^2\ge c^2\\\frac{a}{b+c}\ge\frac{b}{a+c}\ge\frac{c}{a+b}\end{cases}}\)
Áp dụng BĐT Trebysev ta có:
\(a^2\cdot\frac{a}{b+c}+b^2\cdot\frac{b}{a+c}+c^2\cdot\frac{c}{a+b}\ge\frac{a^2+b^2+c^2}{3}\cdot\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)
\(\ge\frac{a^2+b^2+c^2}{3}\cdot\frac{3}{2}\left(nesbitt\right)\)
\(=\frac{a^2+b^2+c^2}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
Ta chứng minh BĐT sau với các số dương:
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Thật vậy, BĐT tương đương: \(\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)
Áp dụng:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) ; \(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\) ; \(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\)
Cộng vế với vế:
\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)
b.
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Rightarrow\dfrac{3}{a}+\dfrac{3}{b}\ge\dfrac{12}{a+b}\) (1)
\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\Rightarrow\dfrac{2}{b}+\dfrac{2}{c}\ge\dfrac{8}{b+c}\) (2)
\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\) (3)
Cộng vế với vế (1); (2) và (3):
\(\dfrac{4}{a}+\dfrac{5}{b}+\dfrac{3}{c}\ge4\left(\dfrac{3}{a+b}+\dfrac{2}{b+c}+\dfrac{1}{c+a}\right)\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
mình nghĩ đề bài sai một chỗ :\(\frac{a^2}{b^2}\)chứ ko phải là \(\frac{a}{b^2}\)
\(\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}=\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\)
=> \(\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Xét: \(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+d^2}+\frac{d^3}{d^2+a^2}\)
\(\Leftrightarrow a-\frac{ab^2}{a^2+b^2}+b-\frac{bc^2}{b^2+c^2}+c-\frac{cd^2}{c^2+d^2}+d-\frac{da^2}{d^2+a^2}\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{\begin{matrix}a^2+b^2\ge2\sqrt{a^2b^2}=2ab\\b^2+c^2\ge2\sqrt{b^2c^2}=2bc\\c^2+d^2\ge2\sqrt{c^2d^2}=2cd\\d^2+a^2\ge2\sqrt{d^2a^2}=2da\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\frac{ab^2}{a^2+b^2}\le\frac{ab^2}{2ab}=\frac{b}{2}\\\frac{bc^2}{b^2+c^2}\le\frac{bc^2}{2bc}=\frac{c}{2}\\\frac{cd^2}{c^2+d^2}\le\frac{cd^2}{2cd}=\frac{d}{2}\\\frac{da^2}{d^2+a^2}\le\frac{da^2}{2da}=\frac{a}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}a-\frac{ab^2}{a^2+b^2}\ge a-\frac{b}{2}\\b-\frac{bc^2}{b^2+c^2}\ge b-\frac{c}{2}\\c-\frac{cd^2}{c^2+d^2}\ge c-\frac{d}{2}\\d-\frac{da^2}{d^2+a^2}\ge d-\frac{a}{2}\end{matrix}\right.\)
\(\Rightarrow a-\frac{ab^2}{a^2+b^2}+b-\frac{bc^2}{b^2+c^2}+c-\frac{cd^2}{c^2+d^2}+d-\frac{da^2}{d^2+a^2}\ge a+b+c+d-\frac{a}{2}-\frac{b}{2}-\frac{c}{2}-\frac{d}{2}\)
\(\Rightarrow a-\frac{ab^2}{a^2+b^2}+b-\frac{bc^2}{b^2+c^2}+c-\frac{cd^2}{c^2+d^2}+d-\frac{da^2}{d^2+a^2}\ge\frac{a+b+c+d}{2}\)
\(\Leftrightarrow\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+d^2}+\frac{d^3}{d^2+a^2}\ge\frac{a+b+c+d}{2}\) ( đpcm )
Cách của bạn Minh dài quá mình xin làm cách ngắn hơn:
Đầu tiên ta chứng minh bổ đề:
\(\frac{x^3}{x^2+y^2}\ge\frac{2x-y}{2}\)
\(\Leftrightarrow2x^3-\left(x^2+y^2\right)\left(2x-y\right)\ge0\)
\(\Leftrightarrow y\left(y-x\right)^2\ge0\)(đúng)
Từ đó ta có: \(\left\{\begin{matrix}\frac{a^3}{a^2+b^2}\ge\frac{2a-b}{2}\\\frac{b^3}{b^2+c^2}\ge\frac{2b-c}{2}\\\frac{c^3}{c^2+d^2}\ge\frac{2c-d}{2}\\\frac{d^3}{d^2+a^2}\ge\frac{2d-a}{2}\end{matrix}\right.\)
Cộng 4 cái trên vế theo vế ta được
\(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+d^2}+\frac{d^3}{d^2+a^2}\ge\frac{2a-b}{2}+\frac{2b-c}{2}+\frac{2c-d}{2}+\frac{2d-a}{2}=\frac{a+b+c+d}{2}\)
By Cauchy-Schwarz, we have:
\(VT\ge\frac{\left(a^3+b^3+c^3\right)^2}{2\left(a^3+b^3+c^3\right)+a^2b+b^2c+c^2a}\)
We will prove: \(a^2b+b^2c+c^2a\le a^3+b^3+c^3\)
\(\Leftrightarrow a^2b+b^2c+c^2a+3abc\le a^3+b^3+c^3+3abc\)
By Schur, we have: \(RHS\ge ab\left(a+b\right)+bc\left(b+c\right)+ca\left(a\right)\)
So we're only need to prove: \(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\ge a^2b+b^2c+c^2a+3abc\)
\(\Leftrightarrow ab^2+bc^2+ca^2\ge3abc\)
It is true by AM-GM ineq', so we have Q.E.D.
P/s: Em thử giải bài này bằng tiếng Anh (để tự luyện kĩ năng tiếng anh, tí em giải lại theo tiếng việt)