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\(\frac{ab}{6+a-c}=\frac{ab}{a+b+c+a-c}=\frac{ab}{2a+b}\)
Áp dụng BĐT Cauchy-schwarz ta có:
\(\frac{ab}{2a+b}\le\frac{ab}{9}.\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)=\frac{2b+a}{9}\)
Chứng minh tương tự ta có:
\(\frac{bc}{2b+c}\le\frac{bc}{9}.\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)=\frac{2c+b}{9}\)
\(\frac{ca}{2c+a}\le\frac{ac}{9}.\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)=\frac{2a+c}{9}\)
Dấu " = " xảy ra <=> a=b=c
Cộng vế với vế của 3 BĐT trên ta có:
\(\frac{ab}{6+a-c}+\frac{bc}{6+b-a}+\frac{ac}{6+c-b}\)
\(=\frac{ab}{2a+b}+\frac{bc}{2b+c}+\frac{ca}{2c+a}\le\frac{3\left(a+b+c\right)}{9}=\frac{6}{3}=2\)
Dấu " = " xảy ra <=> a=b=c=2
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Đặt \(P=a^2+b^2+c^2+ab+bc+ca\)
\(P=\dfrac{1}{2}\left(a+b+c\right)^2+\dfrac{1}{2}\left(a^2+b^2+c^2\right)\)
\(P\ge\dfrac{1}{2}\left(a+b+c\right)^2+\dfrac{1}{6}\left(a+b+c\right)^2=6\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Lời giải:
Đặt biểu thức đã cho là $P$
Do $a+b+c=6$ nên:
$P=\frac{ab}{2a+b}+\frac{bc}{2b+c}+\frac{ca}{2c+a}$
$2P=\frac{2ab}{2a+b}+\frac{2bc}{2b+c}+\frac{2ca}{2c+a}$
$=b-\frac{b^2}{2a+b}+c-\frac{c^2}{2b+c}+a-\frac{a^2}{2c+a}$
$=a+b+c-\left(\frac{b^2}{2a+b}+\frac{c^2}{2b+c}+\frac{a^2}{2c+a}\right)$
Áp dụng BĐT Cauchy-Schwarz:
$\left(\frac{b^2}{2a+b}+\frac{c^2}{2b+c}+\frac{a^2}{2c+a}\right)\geq \frac{(b+c+a)^2}{2a+b+2b+c+2c+a}=\frac{a+b+c}{3}$
Do đó: $2P\leq a+b+c-\frac{a+b+c}{3}=\frac{2}{3}(a+b+c)=\frac{2}{3}.6=4$
$\Rightarrow P\leq 2$ (đpcm)
Dấu "=" xảy ra khi $a=b=c=2$
Lời giải:
Đặt biểu thức đã cho là $P$
Do $a+b+c=6$ nên:
$P=\frac{ab}{2a+b}+\frac{bc}{2b+c}+\frac{ca}{2c+a}$
$2P=\frac{2ab}{2a+b}+\frac{2bc}{2b+c}+\frac{2ca}{2c+a}$
$=b-\frac{b^2}{2a+b}+c-\frac{c^2}{2b+c}+a-\frac{a^2}{2c+a}$
$=a+b+c-\left(\frac{b^2}{2a+b}+\frac{c^2}{2b+c}+\frac{a^2}{2c+a}\right)$
Áp dụng BĐT Cauchy-Schwarz:
$\left(\frac{b^2}{2a+b}+\frac{c^2}{2b+c}+\frac{a^2}{2c+a}\right)\geq \frac{(b+c+a)^2}{2a+b+2b+c+2c+a}=\frac{a+b+c}{3}$
Do đó: $2P\leq a+b+c-\frac{a+b+c}{3}=\frac{2}{3}(a+b+c)=\frac{2}{3}.6=4$
$\Rightarrow P\leq 2$ (đpcm)
Dấu "=" xảy ra khi $a=b=c=2$
\(P=\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\)
\(P\le\dfrac{ab}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\dfrac{bc}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{ca}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\)
\(\Leftrightarrow P\le\dfrac{1}{2}\left(a+b+c\right)=3\)
\(P_{max}=3\) khi \(a=b=c\)
\(a+b+c=7\Rightarrow a+b+c-1=6\)
Ta có:\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Leftrightarrow49=23+2\left(ab+bc+ca\right)\Leftrightarrow ab+bc+ca=13\)
Lại có \(ab+c-6=ab+c-\left(a+b+c-1\right)=ab-a-b+1=\left(a-1\right)\left(b-1\right)\)
Tương tự \(bc+a-6=\left(b-1\right)\left(c-1\right)\)
\(ca+b-6=\left(c-1\right)\left(a-1\right)\)
\(\Rightarrow A=\frac{1}{\left(a-1\right)\left(b-1\right)}+\frac{1}{\left(b-1\right)\left(c-1\right)}+\frac{1}{\left(c-1\right)\left(a-1\right)}\)
\(=\frac{c-1+a-1+b-1}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}=\frac{a+b+c-3}{abc-\left(ab+ac+bc\right)+\left(a+b+c\right)-1}\)
\(=\frac{7-3}{3-13+7-1}=-1\)
\(VP=\frac{6}{\sqrt{\left(3a+bc\right)\left(3b+ca\right)\left(3c+ab\right)}}\)
\(=\frac{6}{\sqrt{\left[\left(a+b+c\right)a+bc\right]\left[\left(a+b+c\right)b+ca\right]\left[\left(a+b+c\right)c+ab\right]}}\)
\(=\frac{6}{\sqrt{\left(a+b\right)^2\left(b+c\right)^2\left(c+1\right)^2}}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)
\(VT=\frac{1}{3a+bc}+\frac{1}{3b+ca}+\frac{1}{3c+ab}\)
\(=\frac{1}{\left(a+b+c\right)a+bc}+\frac{1}{\left(a+b+c\right)b+ac}+\frac{1}{\left(a+b+c\right)c+ab}\)
\(=\frac{\left(b+c\right)+\left(a+c\right)+\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)
Vậy VT = VP, đẳng thức được chứng minh
Áp dụng BĐT Cauchy-Schwarz ta có:
\(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\Rightarrow\left(a+b+c\right)^2\le9\Rightarrow a+b+c\le3\left(1\right)\)
Ta có:\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\ge0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow ab+bc+ca\le3\left(2\right)\)
Cộng vế với vế của\(\left(1\right),\left(2\right)\)ta được:
\(a+b+c+ab+bc+ca\le3+3=6\left(đpcm\right)\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{ab}{6+a-c}=\frac{ab}{a+b+c+a-c}=\frac{ab}{2a+b}\)
\(=\frac{ab}{a+a+b}\le\frac{1}{9}\left(\frac{ab}{a}+\frac{ab}{a}+\frac{ab}{b}\right)=\frac{1}{9}\left(2b+a\right)\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\frac{bc}{6+b-a}\le\frac{1}{9}\left(2c+b\right);\frac{ca}{6+c-b}\le\frac{1}{9}\left(2a+c\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le\frac{1}{9}\cdot3\left(a+b+c\right)=\frac{1}{3}\cdot\left(a+b+c\right)=\frac{6}{3}=2\)
Đẳng thức xảy ra khi \(a=b=c=2\)