cho a, b, c \(\ne\) 0. CMR:
\(\dfrac{a^3-b^3}{ab^2}+\dfrac{b^3-c^3}{bc^2}+\dfrac{c^3-a^3}{ca^2}\ge0\)
K
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27 tháng 12 2018
\(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
\(\Rightarrow VT=\dfrac{ab}{a^2+b^2-c^{^2}}+\dfrac{bc}{b^2+c^2-a^{^2}}+\dfrac{ca}{c^2+a^2-b^{^2}}\\ =\dfrac{ab}{a^2+\left(b+c\right)\left(b-c\right)}+\dfrac{bc}{b^2+\left(c+a\right)\left(c-a\right)}+\dfrac{ca}{c^2+\left(a+b\right)\left(a-b\right)}\\ =\dfrac{ab}{a^2-a\left(b-c\right)}+\dfrac{bc}{b^2-b\left(c-a\right)}+\dfrac{ca}{c^2-c\left(a-b\right)}\\ =\dfrac{b}{a-b+c}+\dfrac{c}{b-c+a}+\dfrac{a}{c-a+b}\\ =\dfrac{b}{\left(a+c\right)-b}+\dfrac{c}{\left(a+b\right)-c}+\dfrac{a}{\left(c+b\right)-a}\\ =\dfrac{b}{-b-b}+\dfrac{c}{-c-c}+\dfrac{a}{-a-a}\\ =\dfrac{b}{-2b}+\dfrac{c}{-2c}+\dfrac{a}{-2a}\\ =-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}=-\dfrac{3}{2}=VP\)
21 tháng 9 2021
Áp dụng BĐT AG-GM:
\(\dfrac{a^3}{a^2+ab+b^2}\ge\dfrac{a^3}{a^2+\dfrac{a^2+b^2}{2}+b^2}=\dfrac{a^3}{\dfrac{3}{2}\left(a^2+b^2\right)}\)
Cmtt \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^3}{b^2+bc+c^2}\ge\dfrac{b^3}{\dfrac{3}{2}\left(b^2+c^2\right)}\\\dfrac{c^3}{c^2+ac+a^2}\ge\dfrac{c^3}{\dfrac{3}{2}\left(c^2+a^2\right)}\end{matrix}\right.\)
Cộng vế theo vế của bất đẳng thức:
\(\Leftrightarrow VT\ge\dfrac{2}{3}\left(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\right)\)
Tiếp tục áp dụng BĐT AG-GM:
\(\dfrac{a^3}{a^2+b^2}=\dfrac{a\left(a^2+b^2\right)-ab^2}{a^2+b^2}=a-\dfrac{ab^2}{a^2+b^2}\ge a-\dfrac{ab^2}{2ab}=a-\dfrac{b}{2}\)
Cmtt\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^3}{b^2+c^2}\ge b-\dfrac{c}{2}\\\dfrac{c^3}{c^2+a^2}\ge c-\dfrac{a}{2}\end{matrix}\right.\)
Cộng vế theo vế
\(\Leftrightarrow VT\ge\dfrac{2}{3}\left(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\right)\\ \ge\dfrac{2}{3}\left(a-\dfrac{b}{2}+b-\dfrac{c}{2}+c-\dfrac{a}{2}\right)=\dfrac{2}{3}\left(a+b+c-\dfrac{a+b+c}{2}\right)=\dfrac{a+b+c}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 9 2021
\(\dfrac{a^3}{a^2+ab+b^2}=a-\dfrac{ab\left(a+b\right)}{a^2+ab+b^2}\ge a-\dfrac{ab\left(a+b\right)}{3\sqrt[3]{a^2.ab.b^2}}=a-\dfrac{a+b}{3}=\dfrac{2a-b}{3}\)
Tương tự và cộng lại ta sẽ có đpcm
HM
cho a,b,c>0 và a+b+c=3
CMR: \(\dfrac{a}{b^3+ab}+\dfrac{b}{c^3+bc}+\dfrac{c}{a^3+ca}\ge\dfrac{3}{2}\)
0
AH
Akai Haruma
Giáo viên
5 tháng 1 2018
Bài 1:
Từ \(a+b+c=0\) ta có:
\(B=\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-b^2-a^2}\)
\(=\frac{a^2}{(-b-c)^2-b^2-c^2}+\frac{b^2}{(-c-a)^2-c^2-a^2}+\frac{c^2}{(-b-a)^2-b^2-a^2}\)
\(=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{a^3+b^3+c^3}{2abc}\)
Lại có:
\(a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3=(-c)^3-3ab(-c)+c^3\)
\(=-c^3+3abc+c^3=3abc\)
Do đó \(B=\frac{3abc}{2abc}=\frac{3}{2}\)
AH
Akai Haruma
Giáo viên
5 tháng 1 2018
Bài 2:
Lấy P-Q ta có:
\(P-Q=\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\right)-\left(\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\right)\)
\(P-Q=\frac{a^3-b^3}{a^2+ab+b^2}+\frac{b^3-c^3}{b^2+bc+c^2}+\frac{c^3-a^3}{c^2+ac+a^2}\)
\(P-Q=\frac{(a-b)(a^2+ab+b^2)}{a^2+ab+b^2}+\frac{(b-c)(b^2+bc+c^2)}{b^2+bc+c^2}+\frac{(c-a)(c^2+ac+a^2)}{c^2+ac+a^2}\)
\(P-Q=(a-b)+(b-c)+(c-a)=0\Rightarrow P=Q\)
Ta có đpcm.
14 tháng 8 2018
Ta có :
\(VT=\dfrac{a^3}{b+c}+\dfrac{b^3}{a+c}+\dfrac{c^3}{a+b}=\dfrac{a^4}{ab+ac}+\dfrac{b^4}{bc+ab}+\dfrac{c^4}{ac+bc}\)
Theo BĐT Cauchy ta có :
\(\dfrac{a^4}{ab+ac}+\dfrac{b^4}{bc+ab}+\dfrac{c^4}{ac+bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ac\right)}\)
Theo BĐT Cô - Si ta lại có : \(a^2+b^2+c^2\ge ab+bc+ac\)
\(\Rightarrow VT\ge\dfrac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}=\dfrac{ab+bc+ca}{2}=\dfrac{1}{2}\)
24 tháng 6 2021
Đặt A = \(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}=0\)
= \(\dfrac{a-b}{c^2+ab+bc+ca}+\dfrac{b-c}{a^2+ab+bc+ca}+\dfrac{c-a}{b^2+ab+bc+ca}\)
= \(\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(c+a\right)}+\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}\)
= \(\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c+a\right)\left(c-a\right)}{\left(c+a\right)\left(b+c\right)\left(a+b\right)}\)
= \(\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
AT
24 tháng 6 2021
\(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}\)
\(=\dfrac{a-b}{ab+bc+ca+c^2}+\dfrac{b-c}{ab+bc+ca+a^2}+\dfrac{c-a}{ab+bc+ca+b^2}\)
\(=\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(a+c\right)}+\dfrac{c-a}{\left(b+a\right)\left(b+c\right)}\)
\(=\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
\(\dfrac{a^3-b^3}{ab^2}+\dfrac{b^3-c^3}{bc^2}+\dfrac{c^3-a^3}{ca^2}\ge0\)
\(\Leftrightarrow\dfrac{a^2}{b^2}-\dfrac{b}{a}+\dfrac{b^2}{c^2}-\dfrac{c}{b}+\dfrac{c^2}{a^2}-\dfrac{a}{c}\ge0\)
Ta có: \(\left\{{}\begin{matrix}\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge\dfrac{2a}{c}\\\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{2b}{a}\\\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge\dfrac{2c}{b}\end{matrix}\right.\)
Cộng 3 cái vế theo vế rồi rút gọn cho 2 ta được ĐPCM
thanks nhiều:))