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\(\frac{a+3c}{a+b}+\frac{a+3b}{a+c}+\frac{2a}{b+c}\)
\(=\frac{a+c+2c}{a+b}+\frac{a+b+2b}{a+c}+\frac{2a}{b+c}\)
\(=\frac{a+c}{a+b}+\frac{2c}{a+b}+\frac{a+b}{a+c}+\frac{2b}{a+c}+\frac{2a}{b+c}\)
\(=\left(\frac{a+c}{a+b}+\frac{a+b}{a+c}\right)+2\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\)
Áp dụng bất đẳng thức AM-GM ta có :
\(\frac{a+c}{a+b}+\frac{a+b}{a+c}\ge2\sqrt{\frac{a+c}{a+b}\cdot\frac{a+b}{a+c}}=2\)
Cần chứng minh \(2\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\ge3\)thì bài toán được chứng minh
tức là \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\frac{3}{2}\)
Tuy nhiên đây là bất đẳng thức Nesbitt quen thuộc nên ta có điều phải chứng minh
Đẳng thức xảy ra <=> a=b=c
Áp dụng BĐT \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) ta có:
\(\frac{1}{a+3b}+\frac{1}{a+b+2c}\ge\frac{4}{2a+4b+2c}=\frac{2}{a+2b+c}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\frac{1}{b+3c}+\frac{1}{2a+b+c}\ge\frac{2}{a+b+2c};\frac{1}{c+3a}+\frac{1}{a+2b+c}\ge\frac{2}{2a+b+c}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT=\frac{1}{b+3c}+\frac{1}{c+3a}+\frac{1}{a+3b}\)
\(\ge\frac{1}{a+b+2c}+\frac{1}{2a+b+c}+\frac{1}{a+2b+c}=VP\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{c+a}\geq \frac{9}{b+c+c+a+c+a}=\frac{9}{3c+2a+b}\)
\(\frac{1}{a+c}+\frac{1}{a+b}+\frac{1}{a+b}\geq \frac{9}{a+c+a+b+a+b}=\frac{9}{3a+2b+c}\)
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{b+c}\geq \frac{9}{a+b+b+c+b+c}=\frac{9}{3b+2c+a}\)
Cộng theo vế rồi rút gọn ta thu được
\(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\geq 3\left(\frac{1}{3a+2b+c}+\frac{1}{3b+2c+a}+\frac{1}{3c+2a+b}\right)\) (đpcm)
Dấu bằng xảy ra khi $a=b=c$
Theo BĐT Bunyakovsky, ta có: \(\frac{7}{2a+b+c}=\frac{7^2}{7\left(2a+b+c\right)}=\frac{\left(2+1+4\right)^2}{2\left(a+3b\right)+\left(b+3c\right)+4\left(c+3a\right)}\)
\(\le\frac{2^2}{2\left(a+3b\right)}+\frac{1^2}{\left(b+3c\right)}+\frac{4^2}{4\left(c+3a\right)}\)
\(=\frac{2}{a+3b}+\frac{1}{b+3c}+\frac{4}{c+3a}\)(1)
Hoàn toàn tương tự: \(\frac{7}{2b+c+a}\le\frac{2}{b+3c}+\frac{1}{c+3a}+\frac{4}{a+3b}\)(2); \(\frac{7}{2c+a+b}\le\frac{2}{c+3a}+\frac{1}{a+3b}+\frac{4}{b+3c}\)(3)
Cộng theo từng vế của 3 BĐT (1), (2), (3), ta được:
\(7\left(\frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b}\right)\le7\left(\frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a}\right)\)
hay \(\frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a}\ge\frac{1}{a+2b+c}+\frac{1}{b+2c+a}+\frac{1}{c+2a+b}\left(q.e.d\right)\)
Đẳng thức xảy ra khi a = b = c
Áp dụng bđt 1/a+1/b >= 4/a+b
Xét 1/a+3b + 1/b+2c+a >= 4/2a+4b+2c = 2/a+2b+c
Tương tự : 1/b+3c + 1/c+2a+b >= 4/2a+2b+4c = 2/a+b+2c
1/c+3a + 1/a+2b+c >= 4/4a+2b+2c = 2/2a+b+c
=> VT + VP >= 2VP
=> VT >= VP ( ĐPCM)
k mk nha
Áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) ta có:
\(\dfrac{1}{a+3b}+\dfrac{1}{a+b+2c}\ge\dfrac{4}{2a+4b+2c}=\dfrac{2}{a+2b+c}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{1}{b+3c}+\dfrac{1}{2a+b+c}\ge\dfrac{2}{a+b+2c};\dfrac{1}{c+3a}+\dfrac{1}{a+2b+c}\ge\dfrac{2}{2a+b+c}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT=\dfrac{1}{b+3c}+\dfrac{1}{c+3a}+\dfrac{1}{a+3b}\)
\(\ge\dfrac{1}{a+b+2c}+\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}=VP\)
\(\frac{ab}{a+3b+2c}=\frac{ab}{a+c+b+c+2b}\le\frac{1}{9}\left(\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{a}{2}\right)\)
Tương tự: \(\frac{bc}{b+3c+2a}\le\frac{1}{9}\left(\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{b}{2}\right)\) ; \(\frac{ca}{c+3a+2b}\le\frac{1}{9}\left(\frac{ca}{b+c}+\frac{ca}{a+b}+\frac{c}{2}\right)\)
Cộng vế với vế:
\(A\le\frac{1}{9}\left(\frac{ab}{a+c}+\frac{bc}{a+c}+\frac{ab}{b+c}+\frac{ca}{b+c}+\frac{bc}{a+b}+\frac{ca}{a+b}+\frac{a+b+c}{2}\right)\)
\(A\le\frac{1}{9}.\frac{3}{2}\left(a+b+c\right)=1\)
Dấu "=" xảy ra khi \(a=b=c=2\)
Áp dụng bất đẳng thức Cauchy-Schwartz ta có
\(\frac{ab}{a+3b+2c}=\frac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\frac{ab}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right).\)
Tương tự ta có 2 bất đẳng thức khác nữa
\(\frac{bc}{b+3c+2a}=\frac{bc}{\left(b+a\right)+\left(a+c\right)+2c}\le\frac{bc}{9}\left(\frac{1}{b+a}+\frac{1}{a+c}+\frac{1}{2c}\right).\)
\(\frac{ac}{c+3a+2b}=\frac{ac}{\left(a+b\right)+\left(b+a\right)+2a}\le\frac{ac}{9}\left(\frac{1}{c+b}+\frac{1}{b+a}+\frac{1}{2a}\right).\)
Cộng ba bất đẳng thức lại cho ta \(\frac{ab}{a+3b+2c}+\frac{bc}{b+3c+2a}+\frac{ca}{c+3a+2b}\le\)
\(\le\frac{ab}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)+\frac{bc}{9}\left(\frac{1}{b+a}+\frac{1}{a+c}+\frac{1}{2c}\right)+\frac{ac}{9}\left(\frac{1}{c+b}+\frac{1}{b+a}+\frac{1}{2a}\right)\)
\(=\frac{1}{9}\left(\frac{ab}{a+c}+\frac{bc}{a+c}\right)+\frac{1}{9}\left(\frac{ab}{b+c}+\frac{ac}{b+c}\right)+\frac{1}{9}\left(\frac{bc}{a+b}+\frac{ac}{a+b}\right)+\frac{a}{18}+\frac{b}{18}+\frac{c}{18}\)
\(=\frac{a+b+c}{6}.\) (ĐPCM)