CMR: (a-b)/(1+ab)+(b-c)/(1+bc)+(c-a)/(1+ac)=3(a-b)/(1+ab).(b-c)/(1+bc).(c-a)/(1+ac)
CMR: \(\frac{a^2+b^2+c^2}{ab+bc+ac} + \frac{1}{3} \geq \frac{8}{9}(\frac{a}{b+c} + \frac{b}{a+c} +\frac{c}{a+b})\)
CMR:\((1+a+b+c)(1+ab+bc+ac) \geq 4\sqrt{2(a+bc)(b+ac)(c+ab)}\)
CMR: a) (a+1)(b+1)(c+1)=abc+ab+ac+bc+a+b+c+1
b) (a-1)(b-1)(c-1)=abc-ab-bc-ac+a+b+c-1
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CMR :
\(\frac{a-b}{1+ab}+\frac{b-c}{1+bc}+\frac{c-a}{1+ac}=\frac{a-b}{1+ab}-\frac{b-c}{1+bc}-\frac{c-a}{1+ac}\)
Là đương nhiên hai biểu thức trên bằng nhau , giống nhau y hệt
a)Cho 3 số dương 0 ≤ a ≤ b ≤ c ≤ 1. CMR : (a/bc+1)+(b/ac+1)+(c/ab+1) ≤ 2
b)Cho a,b,c la 3 canh của 1 Δ. CMR :2(ab+bc+ca) > a2+b2+c2.cho a,b,c duong , a+b+c=1
a, tim Min A=1/(a^2+b^2) +1/(b^2+c^2) +1/(c^2+a^2) +1/ab +1/bc +1/ac
b, tìm Min B=1/(a^2+bc) +1/(b^2+ac) +1/(c^2+ab) +1/ab +1/bc +1/ac
\(a\text{) }\)Áp dụng: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) (a, b > 0). Dấu "=" xảy ra khi a = b.
\(\frac{1}{a^2+b^2}+\frac{1}{ab}=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{1}{2ab}\ge\frac{4}{a^2+b^2+2ab}+\frac{1}{2.\frac{\left(a+b\right)^2}{4}}=\frac{6}{\left(a+b\right)^2}\)
\(=6\left[\frac{1}{\left(a+b\right)^2}+\frac{27}{8}\left(a+b\right)+\frac{27}{8}\left(a+b\right)\right]-\frac{81}{2}\left(a+b\right)\)
\(\ge6.3\sqrt[3]{\frac{1}{\left(a+b\right)^2}.\frac{27}{8}\left(a+b\right).\frac{27}{8}\left(a+b\right)}-\frac{81}{2}\left(a+b\right)\)
\(=\frac{81}{2}-\frac{81}{2}\left(a+b\right)\)
Tương tự: \(\frac{1}{b^2+c^2}+\frac{1}{bc}\ge\frac{81}{2}-\frac{81}{2}\left(b+c\right)\)
\(\frac{1}{c^2+a^2}+\frac{1}{ca}\ge\frac{81}{2}-\frac{81}{2}\left(c+a\right)\)
Cộng theo vế ta được
\(A\ge3.\frac{81}{2}-81\left(a+b+c\right)=3.\frac{81}{2}-81=\frac{81}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}.\)
Vậy GTNN của A là \(\frac{81}{2}.\)
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)^2 = a^2 + b^2 +c^2 và abc khác 0
cmr bc/a^2 + ac/b^2 +ab/c^2 = 3
cho abc=1. rút gọn
a/ab+a+1 + b/bc+b+1 + c/ca+c+1
Cho a,b,c,d thuoc [0,1]. CMR a/(bc+cd+db+1) +b/(cd+da+ac+1) +c(da+ab+bd+1)+d/(ab+bc+ca+1)<= 3/4 +1/4abcd
Cho x>y TM: x+y<=1 CMR: 1/x^2+y^2 = 1/xy>=6
Cho a,b,c >0 TM: a+b+c<=1 CMR: (1/a^2+bc) + (1/b^2+ac)+ 1/c^2+2ab >=9
Cho a,b>0 TM: a+b<=1 ;CMR: (1/a^b^2)+4b+1/ab>=7
Cho a,b>0 TM:a+b<=1. CMR: 1/1+a^2+b^2 +1/2ab >=8/3
Cho a,b,c>0 TM: a+b+c<=3.CMR: 1/a^2+b^2+c^2 +2009/ab+bc+ac >=670