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Áp dụng bất đẳng thức cho hai số dương
\(\dfrac{1}{\left(a+b\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Xét \(c+1=c+a+b+c\)
\(\dfrac{ab}{c+1}\le\dfrac{ab}{4\left[\dfrac{1}{a+c}+\dfrac{1}{b+c}\right]}\)
Tương tự:
\(\dfrac{bc}{a+1}\le\dfrac{bc}{4\left[\dfrac{1}{a+c}+\dfrac{1}{b+a}\right]}\)
\(\dfrac{ca}{b+1}\le\dfrac{ac}{4\left[\dfrac{1}{a+b}+\dfrac{1}{c+b}\right]}\)
Cộng lại :
\(\dfrac{ab}{c+1}+\dfrac{bc}{a+1}+\dfrac{ca}{b+1}\le\dfrac{1}{4}\left[\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{bc}{a+c}+\dfrac{bc}{a+b}+\dfrac{ac}{a+b}+\dfrac{ac}{b+c}\right]\)
Rút gọn mẫu số
\(\Rightarrow\dfrac{ab}{c+1}+\dfrac{bc}{a+1}+\dfrac{ca}{b+1}\le\dfrac{1}{4}\left(a+b+c\right)=\dfrac{1}{4}\)
cho a,b,c khac nhau doi mot va 1/a+1/b+1/c=0.rut gon cac bieu thuc
N=bc/a^2+2bc+CA/B^2+2AC+AB/C^2+2AB
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thay ab+bc+ac=1 vào 1+a^2=ab+bc+ca+a^2=b*(a+c)+a*( a+c)=(a+b)*(a+c)
tương tự 1+b^2=(a+b)*(b+c);1+c^2=(a+c)*(b+c)
mẫu số của A=(a+b)^2*(b+c)^2*(c+a)^2=Tử số của A
=> A=1
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Áp dụng tính chất : 1/x+y < = 1/4.(1/x + 1/y) với x,y > 0 thì :
ab/c+1 = ab/c+a+b+c = ab/(c+a)+(c+b) < = ab/4.(1/c+a + 1/c+b) = 1/4.(ab/c+a + ab/c+b)
Tương tự : bc/a+1 < = 1/4.(bc/a+c + bc/a+b) ; ca/b+a < = 1/4.(ca/b+c + ca/b+a)
=> ab/c+1 + bc/a+1 + ca/b+1 < = 1/4.(ab/c+a + ab/c+b + bc/a+c + bc/a+b + ca/b+c + ca/b+a )
= 1/4.[(ab/c+a + bc/a+c) + (ab/c+b + ca/b+c) + (bc/a+b + ca/a+b)]
= 1/4.(a+b+c) = 1/4
=> ĐPCM
Tk mk nha
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\(a\text{) }\)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 \(\frac{81}{2}.\)