Cho A = \(\dfrac{b^2}{a+b}+\dfrac{c^2}{b+c}+\dfrac{a^2}{a+c}\); B= \(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{a+c}\). CMR A = B
K
Khách
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Những câu hỏi liên quan
3 tháng 1 2019
3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
AH
Akai Haruma
Giáo viên
4 tháng 1 2019
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
19 tháng 2 2021
a) Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{b}{a}=\dfrac{d}{c}\)
\(\Leftrightarrow\dfrac{b}{a}-1=\dfrac{d}{c}-1\)
\(\Leftrightarrow\dfrac{b-a}{a}=\dfrac{d-c}{c}\)
\(\Leftrightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
\(\Leftrightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\)(đpcm)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 5 2021
a.
Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge2\sqrt{\dfrac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\)
Tương tự: \(\dfrac{b^2}{c+a}+\dfrac{c+a}{4}\ge b\) ; \(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)
Cộng vế:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{a+b+c}{2}\ge a+b+c\)
\(\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 5 2021
b.
Ta có:
\(a^2+bc\ge2\sqrt{a^2bc}=2\sqrt{ab.ac}\Rightarrow\dfrac{1}{a^2+bc}\le\dfrac{1}{2\sqrt{ab.ac}}\le\dfrac{1}{4}\left(\dfrac{1}{ab}+\dfrac{1}{ac}\right)\)
Tương tự: \(\dfrac{1}{b^2+ac}\le\dfrac{1}{4}\left(\dfrac{1}{ab}+\dfrac{1}{bc}\right)\) ; \(\dfrac{1}{c^2+ab}\le\dfrac{1}{4}\left(\dfrac{1}{ac}+\dfrac{1}{bc}\right)\)
Cộng vế với vế:
\(\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{1}{2}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\dfrac{a+b+c}{2abc}\)
Dấu "=" xảy ra khi \(a=b=c\)
HP
30 tháng 1 2021
1.
Áp dụng BĐT BSC:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
Đẳng thức xảy ra khi \(a=b=c>0\)
2.
Áp dụng BĐT \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\) và BĐT BSC:
\(\dfrac{a+b}{a^2+b^2}+\dfrac{b+c}{b^2+c^2}+\dfrac{c+a}{c^2+a^2}\)
\(\le\dfrac{a+b}{\dfrac{\left(a+b\right)^2}{2}}+\dfrac{b+c}{\dfrac{\left(b+c\right)^2}{2}}+\dfrac{c+a}{\dfrac{\left(c+a\right)^2}{2}}\)
\(=\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)
\(\le2.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}\right)\)
\(=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Đẳng thức xảy ra khi \(a=b=c>0\)
HP
30 tháng 1 2021
Cách khác:
1.
Áp dụng BĐT Cauchy:
\(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}+\dfrac{b^2}{c+a}+\dfrac{c+a}{4}+\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge a+b+c\)
\(\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge a+b+c-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}\)
Đẳng thức xảy ra khi \(a=b=c>0\)
HP
24 tháng 2 2021
Hình như thế này mới đúng chứ \(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\)
Áp dụng BĐT Cosi:
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2.\dfrac{a}{c};\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2.\dfrac{b}{a};\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2.\dfrac{c}{b}\)
\(\Rightarrow2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\)
Đẳng thức xảy ra khi \(a=b=c>0\)
YN
6 tháng 2 2023
a) Đề là chứng minh \(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\) à bạn?
Ta có: \(\dfrac{a}{c}=\dfrac{c}{b}\)
\(\Rightarrow ab=c^2\)
\(\Rightarrow\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a^2+ab}{b^2+ab}=\dfrac{a\left(a+b\right)}{b\left(a+b\right)}=\dfrac{a}{b}\)
\(\Rightarrowđpcm\)
b)
Ta có: \(\dfrac{a}{c}=\dfrac{c}{d}\)
\(\Rightarrow c^2=ab\)
\(\Rightarrow\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b^2-a^2}{a^2+ab}=\dfrac{\left(b-a\right)\left(b+a\right)}{a\left(a+b\right)}=\dfrac{b-a}{a}\)
\(\Rightarrowđpcm.\)
22 tháng 2 2018
a, Từ \(\dfrac{a}{c}=\dfrac{c}{b}\Rightarrow c^2=a\cdot b\). Khi đó :
\(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a^2+a\cdot b}{b^2+a\cdot b}=\dfrac{a\cdot\left(a+b\right)}{b\cdot\left(a+b\right)}=\dfrac{a}{b}=VP\)
⇒ĐPCM
b, Từ \(\dfrac{a}{c}=\dfrac{c}{b}\Rightarrow c^2=a\cdot b\) và áp dụng công thức \(a^2-b^2=\left(a+b\right)\cdot\left(a-b\right)\).Khi đó :
\(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{\left(b-a\right)\left(b+a\right)}{a^2+a\cdot b}=\dfrac{\left(b-a\right)\left(a+b\right)}{a\cdot\left(a+b\right)}=\dfrac{b-a}{a}=VP\)
⇒ĐPCM
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 3 2021
\(\Leftrightarrow\dfrac{2a^2}{b^2}+\dfrac{2b^2}{c^2}+\dfrac{2c^2}{a^2}=\dfrac{2a}{c}+\dfrac{2c}{b}+\dfrac{2b}{a}\)
\(\Leftrightarrow\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}-\dfrac{2a}{c}\right)+\left(\dfrac{a^2}{b^2}+\dfrac{c^2}{a^2}-\dfrac{2c}{b}\right)+\left(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}-\dfrac{2b}{a}\right)=0\)
\(\Leftrightarrow\left(\dfrac{a}{b}-\dfrac{b}{c}\right)^2+\left(\dfrac{a}{b}-\dfrac{c}{a}\right)^2+\left(\dfrac{b}{c}-\dfrac{c}{a}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}-\dfrac{b}{c}=0\\\dfrac{a}{b}-\dfrac{c}{a}=0\\\dfrac{b}{c}-\dfrac{c}{a}=0\end{matrix}\right.\) \(\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\Leftrightarrow a=b=c\)