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Áp dụng BĐT AM-GM ta có:
\(\dfrac{1}{1-ab}=1+\dfrac{ab}{1-ab}\le1+\dfrac{ab}{1-\dfrac{a^2+b^2}{2}}=1+\dfrac{2ab}{a^2+b^2+2c^2}\)
\(=1+\dfrac{2ab}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}\le1+\dfrac{ab}{\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}}\)
\(\le1+\dfrac{1}{2}\left(\dfrac{a^2}{a^2+c^2}+\dfrac{b^2}{b^2+c^2}\right)\). Tương tự ta cũng có:
\(\dfrac{1}{1-bc}\le1+\dfrac{1}{2}\left(\dfrac{b^2}{a^2+b^2}+\dfrac{c^2}{a^2+c^2}\right);\dfrac{1}{1-ca}\le1+\dfrac{1}{2}\left(\dfrac{c^2}{b^2+c^2}+\dfrac{a^2}{a^2+b^2}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le3+\dfrac{1}{2}\left(\dfrac{a^2+b^2}{a^2+b^2}+\dfrac{b^2+c^2}{b^2+c^2}+\dfrac{c^2+a^2}{c^2+a^2}\right)=\dfrac{9}{2}\)
Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
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:\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
\(\Rightarrow A\le a+b+c-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}\)
\(\Rightarrowđpcm\)
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:
\(VT\le a+b+c-\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=a+b+c-\dfrac{a+b+c}{2}=\dfrac{1}{2}\left(a+b+c\right)\)
Bài 1:
Từ \(a+b+c=0\) ta có:
\(B=\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-b^2-a^2}\)
\(=\frac{a^2}{(-b-c)^2-b^2-c^2}+\frac{b^2}{(-c-a)^2-c^2-a^2}+\frac{c^2}{(-b-a)^2-b^2-a^2}\)
\(=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{a^3+b^3+c^3}{2abc}\)
Lại có:
\(a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3=(-c)^3-3ab(-c)+c^3\)
\(=-c^3+3abc+c^3=3abc\)
Do đó \(B=\frac{3abc}{2abc}=\frac{3}{2}\)
Bài 2:
Lấy P-Q ta có:
\(P-Q=\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\right)-\left(\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\right)\)
\(P-Q=\frac{a^3-b^3}{a^2+ab+b^2}+\frac{b^3-c^3}{b^2+bc+c^2}+\frac{c^3-a^3}{c^2+ac+a^2}\)
\(P-Q=\frac{(a-b)(a^2+ab+b^2)}{a^2+ab+b^2}+\frac{(b-c)(b^2+bc+c^2)}{b^2+bc+c^2}+\frac{(c-a)(c^2+ac+a^2)}{c^2+ac+a^2}\)
\(P-Q=(a-b)+(b-c)+(c-a)=0\Rightarrow P=Q\)
Ta có đpcm.
cho ba số dương \(0\le a\le b\le c\le1\) CMR \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le2\)
Vì \(0\le a\le b\le c\le1\) nên:
\(\left(a-1\right)\left(b-1\right)\ge ab+1\ge a+b\Leftrightarrow\dfrac{1}{ab+1}\le\dfrac{1}{a+b}\Leftrightarrow\dfrac{c}{ab+1}\le\dfrac{c}{a+b}\left(1\right)\)
Tương tự: \(\dfrac{a}{bc+1}\le\dfrac{a}{b=c}\left(2\right);\dfrac{b}{ac+1}\le\dfrac{b}{a+c}\left(3\right)\)
Do đó: \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\left(4\right)\)
Mà: \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\le\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\left(5\right)\)
Từ (4) và (5) suy ra \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\left(đpcm\right)\)
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: \(a=b=c=1\)
Ta có : \(a^2+b^2\ge2ab\Rightarrow a^2+b^2-ab\ge ab\)
\(\Rightarrow\dfrac{1}{a^2-ab+b^2}\le\dfrac{1}{ab}=\dfrac{abc}{ab}=c\) ( do $abc=1$ )
Tương tự ta có :
\(\dfrac{1}{b^2-bc+c^2}\le a\)
\(\dfrac{1}{c^2-ab+a^2}\le b\)
Cộng vế với vế các BĐT trên có :
\(\dfrac{1}{a^2-ab+b^2}+\dfrac{1}{b^2-bc+c^2}+\dfrac{1}{c^2-ac+a^2}\le a+b+c\)
Dấu "=" xảy ra khi $a=b=c$
\(VT=\dfrac{1}{a^2+b^2-ab}+\dfrac{1}{b^2+c^2-bc}+\dfrac{1}{c^2+a^2-ca}\)
\(VT\le\dfrac{1}{2ab-ab}+\dfrac{1}{2bc-bc}+\dfrac{1}{2ca-ca}=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=\dfrac{a+b+c}{abc}=a+b+c\)
Dấu "=" xảy ra khi \(a=b=c=1\)