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a)Svac-so:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{b+c+c+a+a+b}=\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2\left(đpcm\right)}\)
b)\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\ge\dfrac{2}{ab+1}\)
\(\Leftrightarrow\dfrac{1}{a^2+1}-\dfrac{1}{ab+1}+\dfrac{1}{b^2+1}-\dfrac{1}{ab+1}\ge0\)
\(\Leftrightarrow\dfrac{ab+1-a^2-1}{\left(a^2+1\right)\left(ab+1\right)}+\dfrac{ab+1-b^2-1}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\dfrac{a\left(b-a\right)}{\left(a^2+1\right)\left(ab+1\right)}+\dfrac{b\left(a-b\right)}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{b}{\left(b^2+1\right)\left(ab+1\right)}-\dfrac{a}{\left(a^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{b\left(a^2+1\right)-a\left(b^2+1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{a^2b+b-ab^2-a}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{ab\left(a-b\right)-\left(a-b\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\cdot\dfrac{ab-1}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)(luôn đúng)
Áp dụng BĐT Cô - Si , ta có :
\(\dfrac{a}{b^2}+\dfrac{1}{a}\) ≥ \(2\sqrt{\dfrac{a}{b^2}.\dfrac{1}{a}}=2.\dfrac{1}{b}\left(a,b>0\right)\left(1\right)\)
\(\dfrac{b}{c^2}+\dfrac{1}{b}\text{ ≥ }2\sqrt{\dfrac{b}{c^2}.\dfrac{1}{b}}=2.\dfrac{1}{c}\left(b,c>0\right)\left(2\right)\)
\(\dfrac{c}{a^2}+\dfrac{1}{c}\text{≥}2\sqrt{\dfrac{c}{a^2}.\dfrac{1}{c}}=2.\dfrac{1}{a}\left(a,c>0\right)\left(3\right)\)
Từ ( 1 ; 2 ; 3) Ta có :
\(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) ≥ \(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
⇔\(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\) ≥ \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
e)
\(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) ( luôn đúng)
=> ĐPCM
b. Áp dụng BĐT Cauchy-Schwarz dạng Engel, ta có:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=\dfrac{\left(a+b+c\right)^2}{b+c+a+c+a+b}=\dfrac{\left(a+b+c\right)^2}{2a+2b+2c}=\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
a) Áp dụng BĐT Cauchy :
Ta có: \(x+\dfrac{1}{x}\) = \(\dfrac{x^2+1}{x}\) \(\ge\) \(\dfrac{2x}{x}\) = 2 => đpcm
=> Dấu = xảy ra khi x = 1
b) Áp dụng BĐT Svac-sơ ta có:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\) \(\ge\) \(\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
=> đpcm
=> Dấu bằng xảy ra <=> a = b = c
ko xoắn 1 dòng thôi
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{b+c+c+a+a+b}=\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
Áp dụng bđt AM-GM: \(\dfrac{a}{b^2}+\dfrac{1}{a}\ge2\sqrt{\dfrac{a}{b^2a}}=2\sqrt{\dfrac{1}{b^2}}=\dfrac{2}{b}\) \(\dfrac{b}{c^2}+\dfrac{1}{b}\ge2\sqrt{\dfrac{b}{c^2b}}=2\sqrt{\dfrac{1}{c^2}}=\dfrac{2}{c}\) \(\dfrac{c}{a^2}+\dfrac{1}{c}\ge2\sqrt{\dfrac{c}{a^2c}}=2\sqrt{\dfrac{1}{a^2}}=\dfrac{2}{a}\) Cộng theo vế: \(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\Leftrightarrow\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)Dấu "=" xảy ra khi: \(a=b=c\)
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:\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
\(\Rightarrow A\le a+b+c-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}\)
\(\Rightarrowđpcm\)
Áp dụng BĐT Cô si dạng phân số ta có :
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
=> ĐPCM .
b) Vì a,b,c > 0 .
Áp dụng BĐT Cô si ta có :
\(\dfrac{a^2}{b}+b\ge2a\) (1)
Tương tự ta có : \(\dfrac{b^2}{c}+c\ge2b\) (2)
\(\dfrac{c^2}{a}+a\ge2c\) (3)
Cộng từng vế => ĐPCM .