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Áp dụng BĐT Cauchy ta có
\(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge a\)
\(\dfrac{b^2}{a+c}+\dfrac{a+c}{4}\ge b\)
\(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}+\dfrac{a+b+c}{2}\ge a+b+c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)
Dấu bằng xảy ra khi a=b=c
Làm tắt vài chỗ thông cảm
Câu b,
Ta có BĐT Cauchy \(a^2+b^2\ge2ab\)
\(\Rightarrow\left(a+b\right)^2\ge4ab\)
\(\Rightarrow ab\le\dfrac{\left(a+b\right)^2}{4}\)
\(\Rightarrow\dfrac{ab}{a+b}\le\dfrac{\left(a+b\right)^2}{4\left(a+b\right)}=\dfrac{a+b}{4}\)
Tương tự \(\dfrac{bc}{b+c}\le\dfrac{b+c}{4}\)
\(\dfrac{ac}{a+c}\le\dfrac{a+c}{4}\)
Cộng theo vế ta đc \(VT\le\dfrac{2\left(a+b+c\right)}{4}=\dfrac{a+b+c}{2}\)
Dấu bằng xảy ra khi a=b=c
Ta có:\(\dfrac{ab}{a+b}=\dfrac{ab+b^2-b^2}{a+b}=\dfrac{b\left(a+b\right)-b^2}{a+b}=b-\dfrac{b^2}{a+b}\)
Tương tự với các vế ta được:
\(\dfrac{bc}{b+c}=c-\dfrac{c^2}{b+c}\) và \(\dfrac{ac}{a+c}=a-\dfrac{a^2}{a+c}\)
Cộng theo vế:
\(VT=a+b+c-\left(\dfrac{b^2}{a+b}+\dfrac{c^2}{b+c}+\dfrac{a^2}{a+c}\right)\)
Áp dụng bất đẳng thức Cauchy-Schwarz:
\(VT\le a+b+c-\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=a+b+c-\dfrac{a+b+c}{2}=\dfrac{1}{2}\left(a+b+c\right)\)
Ta có:
\(\dfrac{bc}{a}+\dfrac{ac}{b}=c\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge2c\)
Chứng minh tương tự, ta có:
\(\dfrac{bc}{a}+\dfrac{ab}{c}\ge2b\)
\(\dfrac{ac}{b}+\dfrac{ab}{c}\ge2a\)
\(\Rightarrow2\left(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\right)\ge2\left(a+b+c\right)\)
\(\Rightarrow\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\ge a+b+c\)
Dấu = xảy ra khi a = b = c
Áp dụng BĐT Cô - Si cho các số dương , ta có :
\(\dfrac{ab}{c}+\dfrac{bc}{a}\ge2\sqrt{\dfrac{ab}{c}.\dfrac{bc}{a}}=2\sqrt{b^2}=2b\) ( 1)
\(\dfrac{bc}{a}+\dfrac{ac}{b}\ge2\sqrt{\dfrac{bc}{a}.\dfrac{ac}{b}}=2\sqrt{c^2}=2c\) ( 2)
\(\dfrac{ab}{c}+\dfrac{ac}{b}\ge2\sqrt{\dfrac{ab}{c}.\dfrac{ac}{b}}=2\sqrt{a^2}=2a\) ( 3)
Cộng từng vế của ( 1;2;3) , ta có :
\(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\)
Đẳng thức xảy ra khi : a = b = c
Áp dụng bđt cosi ta có:
\(\dfrac{ab}{c}+\dfrac{bc}{a}\ge2\sqrt{\dfrac{ab}{c}\cdot\dfrac{bc}{a}}=2\sqrt{b^2}=2b\)
Tương tự:
\(\left\{{}\begin{matrix}\dfrac{bc}{a}+\dfrac{ac}{b}\ge2b\\\dfrac{ab}{c}+\dfrac{ac}{b}\ge2a\end{matrix}\right.\)
Cộng 2 vế của các bđt trên ta có:
\(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}+\dfrac{ac}{b}\ge2b+2c+2a\)
\(\Rightarrow2\left(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\right)\ge2\left(a+b+c\right)\)
\(\Rightarrow\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\)
Dấu ''='' xảy ra khi a = b = c
Áp dụng bất đẳng thức Cauchy-Schwarz:
\(A=\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ca}\)
\(A\ge\dfrac{\left(1+1+1\right)^2}{3+ab+bc+ac}=\dfrac{9}{3+ab+bc+ac}\)
Mặt khác,theo hệ quả AM-GM: \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}\le\dfrac{3^2}{3}=3\)
\(\Rightarrow\dfrac{9}{3+ab+bc+ac}\ge\dfrac{9}{3+3}=\dfrac{9}{6}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi: \(a=b=c=1\)
a) Áp dụng bất đẳng thức AM-GM ta có:
\(\dfrac{bc}{a}+\dfrac{ac}{b}\ge2\sqrt{\dfrac{abc^2}{ab}}=2\sqrt{c^2}=2\left|c\right|=2c\left(c>0\right)\)
Chứng minh tương tự ta được: \(\left\{{}\begin{matrix}\dfrac{ac}{b}+\dfrac{ab}{c}\ge2a\\\dfrac{bc}{a}+\dfrac{ab}{c}\ge2b\end{matrix}\right.\)
Cộng theo vế: \(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\ge a+b+c\left(đpcm\right)\)
Áp dụng liên tiếp AM-GM và Cauchy-Schwarz ta được:
\(\dfrac{ab}{a+b}=\dfrac{ab+b^2-b^2}{a+b}=\dfrac{b\left(a+b\right)}{a+b}-\dfrac{b^2}{a+b}=b-\dfrac{b^2}{a+b}\)
Chứng minh tương tự:
\(\left\{{}\begin{matrix}\dfrac{bc}{b+c}=\dfrac{bc+c^2-c^2}{b+c}=\dfrac{c\left(b+c\right)}{b+c}-\dfrac{c^2}{b+c}=c-\dfrac{c^2}{b+c}\\\dfrac{ac}{c+a}=\dfrac{ac+a^2-a^2}{c+a}=\dfrac{a\left(c+a\right)}{c+a}-\dfrac{a^2}{c+a}=a-\dfrac{a^2}{c+a}\end{matrix}\right.\)
Cộng theo vế:
\(\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ac}{a+c}=a+b+c-\left(\dfrac{b^2}{a+b}+\dfrac{c^2}{b+c}+\dfrac{a^2}{a+c}\right)\le\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=a+b+c-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}\left(đpcm\right)\)
b)Đặt \(A=\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\)
\(A=\dfrac{a\left(a+b\right)-a^2}{a+b}+\dfrac{b\left(b+c\right)-b^2}{a+b}+\dfrac{c\left(c+a\right)-c^2}{c+a}\)
\(A=a+b+c-\dfrac{a^2}{a+b}-\dfrac{b^2}{b+c}-\dfrac{c^2}{c+a}\)
Lại có:\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
\(\Rightarrow A\le a+b+c-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}\)
\(\Rightarrowđpcm\)