Cho biết: \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)
C/m: M=\(\frac{ab+bc+ca}{a^2+b^2+c}\)= 1
cho a,b,c là các số thực không âm thỏa mãn ab+bc+ca>0. Chứng minh rằng
\(\frac{1}{2a^2+bc}+\frac{1}{2b^2+ca}+\frac{1}{2c^2+ab}+\frac{1}{ab+bc+ca}\ge\frac{12}{\left(a+b+c\right)^2}\)
Em chỉ giải ra được 1 TH dấu bằng thôi: a = b = c (còn trường hợp a = b; c=0 và các hoán vị thì em chịu, vì khi xét dấu = trong bđt thì em chỉ xảy ra 1 th)
Áp dụng BĐT Cauchy-Schwarz dạng Engel;
\(VT\ge\frac{16}{a^2+b^2+c^2+\left(a+b+c\right)^2}\ge\frac{16}{\frac{\left(a+b+c\right)^2}{3}+\left(a+b+c\right)^2}\)\(=\frac{12}{\left(a+b+c\right)^2}\) (đpcm)
Đẳng thức xảy ra khi a = b = c
cho M =\(\frac{b-c}{a^2-ac-ab+bc}+\frac{c-a}{b^2-ab-cb+ca}+\frac{a-b}{c^2-bc-ac+ab}\) và N=\(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\) cmr M=2N
\(M=\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-a\right)}\)
Đánh giá đại diện: \(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{\left(a-c\right)-\left(a-b\right)}{\left(a-b\right)\left(a-c\right)}=\frac{1}{a-b}-\frac{1}{a-c}\)
Tương tự: \(\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{b-c}-\frac{1}{b-a}\)
\(\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{c-a}-\frac{1}{c-b}\)
\(\Rightarrow M=\frac{1}{a-b}-\frac{1}{a-c}+\frac{1}{b-c}-\frac{1}{b-a}+\frac{1}{c-a}-\frac{1}{c-b}\)
\(\Rightarrow M=\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}+\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}\)
\(\Rightarrow M=2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)=2N\left(đpcm\right)\)
cho a,b,c > 0 thỏa mãn ab+bc+ca=1. Cmr:
\(a+b+c+\frac{ab}{b+c}+\frac{bc}{c+a}+\frac{ca}{a+b}\ge\frac{3\sqrt{3}}{2}\)
Lời giải:
Ta thấy:
\(\text{VT}=(a+\frac{ca}{a+b})+(b+\frac{ab}{b+c})+(c+\frac{bc}{c+a})\)
\(=\frac{a(a+b+c)}{a+b}+\frac{b(a+b+c)}{b+c}+\frac{c(a+b+c)}{c+a}\)
\(=(a+b+c)\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)\)
\(\geq (a+b+c).\frac{(a+b+c)^2}{a^2+ab+b^2+bc+c^2+ac}=\frac{(a+b+c)^3}{a^2+b^2+c^2+ab+bc+ac}\) (theo BĐT Cauchy-Schwarz)
Có:
$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=a^2+b^2+c^2+2$
$\Rightarrow a+b+c=\sqrt{a^2+b^2+c^2+2}=\sqrt{t+2}$ với $t=a^2+b^2+c^2$
Do đó:
$\text{VT}\geq \frac{\sqrt{(t+2)^3}}{t+1}$ \(=\sqrt{\frac{(t+2)^3}{(t+1)^2}}\)
Áp dụng BĐT AM-GM:
\((t+2)^3=\left(\frac{t+1}{2}+\frac{t+1}{2}+1\right)^3\geq 27.\frac{(t+1)^2}{4}\)
\(\Rightarrow \text{VT}=\sqrt{\frac{(t+2)^3}{(t+1)^2}}\geq \sqrt{\frac{27}{4}}=\frac{3\sqrt{3}}{2}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=\frac{1}{\sqrt{3}}$
Quay lại diễn đàn trong thinh lặng:))
Chứng minh: $$\left( a+{\frac {ab}{b+c}}+b+{\frac {bc}{c+a}}+c+{\frac {ca}{a+b}}
\right) ^{2}-{\frac {27\,ab}{4}}-{\frac {27\,ca}{4}} \geqq {\frac {27\,bc}{
4}}$$
Sau khi quy đồng, cần chứng minh$:$
$$\frac{1}{2} \sum\limits_{cyc} \left( 5\,{a}^{4}{b}^{2}+8\,{a}^{3}{b}^{3}+7\,{a}^{2}{b}^{4}+98\,{a}^
{2}{b}^{3}c+99\,{a}^{2}{b}^{2}{c}^{2}+124\,{a}^{2}b{c}^{3}+34\,a{b}^{4
}c+130\,a{b}^{3}{c}^{2}+26\,{b}^{4}{c}^{2}+44\,{b}^{3}{c}^{3}+{c}^{6}
\right) \left( a-b \right) ^{2} \geqq 0$$
cho a,b,c và \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)
tính M= \(\frac{ab+bc+ca}{a^2+b^2+c^2}\)
Từ M=\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)
\(\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)
\(\Rightarrow\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{c}{ca}+\frac{a}{ca}\)
\(\Rightarrow\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\)
\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)
\(\Rightarrow a=b=c\)
Ta có: \(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)
Vậy M= 1
Ta có \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{a+c}\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{a+c}{ac}\)
\(\Rightarrow\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{c}{ca}+\frac{a}{ca}\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\)
Có \(\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}\Leftrightarrow\frac{1}{a}=\frac{1}{c}\left(1\right)\) và \(\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\Leftrightarrow\frac{1}{b}=\frac{1}{c}\left(2\right)\)
Từ \(\left(1\right)\left(2\right)\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\) hay \(a=b=c\)
Vậy \(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\)
Cho a,b,c>0. Cmr: a) \(\frac{ab}{a^2+bc+ca}+\frac{bc}{b^2+ca+ab}+\frac{ca}{c^2+ab+bc}\le\frac{a^2+b^2+c^2}{ab+bc+ca}\)
b) \(\frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}\le1\)
a)\(VT=\sum_{cyc}\frac{ab^3+ab^2c+a^2bc}{\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)}\le\frac{\sum_{cyc}\left(ab^3+ab^2c+a^2bc\right)}{\left(ab+bc+ca\right)^2}\)
\(=\frac{ab^3+bc^3+ca^3+2a^2bc+2ab^2c+2abc^2}{\left(ab+bc+ca\right)^2}\)\(\le\frac{\sum_{cyc}ab\left(a^2+b^2\right)+abc\left(a+b+c\right)}{\left(ab+bc+ca\right)^2}\)
\(=\frac{\left(ab+bc+ca\right)\left(a^2+b^2+c^2\right)}{\left(ab+bc+ca\right)^2}=\frac{a^2+b^2+c^2}{ab+bc+ca}=VP\)
Cho a, b, c khác 0 thỏa mãn : \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\) . Tính M=\(\frac{ab+bc+ca}{a^2+b^2+c^2}\)
Từ \(\frac{ab}{a+b}=\frac{bc}{b+c}\Leftrightarrow\frac{abc}{ac+bc}=\frac{abc}{ab+ac}\Leftrightarrow bc=ab\Rightarrow a=c\)(1)
Tương tựi ta cũng có : \(\hept{\begin{cases}a=b\\b=c\end{cases}}\)(2)
Từ (1);(2) \(\Rightarrow a=b=c\)Thay vào M ta được :\(M=\frac{a.a+a.a+a.a}{a^2+b^2+c^2}=1\)
Cho a,b,c khác 0 thỏa mãn: \(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\)
Tính: M=\(\frac{ab+bc+ca}{a^2+b^2+c^2}\)
Cho các số thực dương a,b,c .
Tìm giá trị lớn nhất của biểu thức \(P=\frac{ab}{a^2+ab+bc}+\frac{bc}{b^2+bc+ca}+\frac{ca}{c^2+ca+ab}\)
Cauchy-SChwarz:
\(VT=\sum_{cyc}\frac{ab}{a^2+ab+bc}\le\frac{\sum_{cyc}\left(a^2b^2+ab^2c+abc^2\right)}{\left(ab+bc+ca\right)^2}=\frac{\left(ab+bc+ca\right)^2}{\left(ab+bc+ca\right)^2}=1\)
Dau "=" khi a=b=c\(\in R^+\)
Cho a , b , c > 0 thỏa mãn \(a^2b+b^2c+c^2a=3\)
Chứng minh \(\frac{ab+bc+ca}{2\left(a^2+b^2+c^2\right)}+\frac{1}{6}\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\ge\frac{a+b+c}{3}\)