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\(P=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2}\) (BĐT Cauchy Schwarz)
\(=\dfrac{9}{ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2}\)
\(=\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2}+\dfrac{7}{ab+bc+ca}\)
\(\ge\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2ac+2bc}+\dfrac{7}{ab+bc+ca}\)
\(=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}\)
Ta có: \(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1}{3}\) .Thế vào biểu thức
\(\Rightarrow P\ge9+\dfrac{7}{\dfrac{1}{3}}=9+21=30\)
\(\Rightarrow P_{min}=30\) khi \(a=b=c=\dfrac{1}{3}\)
\(\)Ta có: \(a+b+c=0 \Rightarrow b+c=-a \Rightarrow (b+c)^2=(-a)^2 \Leftrightarrow b^2+c^2+2bc=a^2 \Leftrightarrow a^2-b^2-c^2=2bc\)
Tương tự: \(b^2-c^2-a^2=2ca;c^2-a^2-b^2=2ab\)
\(P=...=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)
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Bổ đề \(a+b+c=0 \Leftrightarrow a^3+b^3+c^3\)
Ở đây ta c/m chiều thuận:
Với \(a+b+c=0 \Leftrightarrow a+b=-c \Rightarrow (a+b)^3=(-c)^3 \Leftrightarrow a^3+b^3+3ab(a+b)=-c^3 \Leftrightarrow a^3+b^3+c^3=3abc(QED)\)
BĐT bị ngược dấu, BĐT đúng phải là:
\(\dfrac{a}{ac+4}+\dfrac{b}{ab+4}+\dfrac{c}{bc+4}\le\dfrac{a^2+b^2+c^2}{16}\)
\(ab+bc+ca\le1\)
\(\Rightarrow\sqrt{a^2+1}\ge\sqrt{a^2+ab+bc+ca}=\sqrt{\left(a+b\right)\left(a+c\right)}\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+1}}\le\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{a}{a+b}+\dfrac{a}{a+c}}{2}\)
\(tương\) \(tự\Rightarrow\Sigma\dfrac{a}{\sqrt{a^2+1}}\le\dfrac{\dfrac{a}{a+b}+\dfrac{a}{a+c}}{2}+\dfrac{\dfrac{b}{a+b}+\dfrac{b}{b+c}}{2}+\dfrac{\dfrac{c}{b+c}+\dfrac{c}{a+c}}{2}=\dfrac{3}{2}\left(đpcm\right)\)
\(dấu"="\Leftrightarrow a=b=c=\sqrt{\dfrac{1}{3}}\)
\(a^2+b^2-ab\ge\dfrac{1}{2}\left(a+b\right)^2-\dfrac{1}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\dfrac{1}{\sqrt{a^2-ab+b^2}}\le\dfrac{1}{\sqrt{\dfrac{1}{4}\left(a+b\right)^2}}=\dfrac{2}{a+b}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tương tự:
\(\dfrac{1}{\sqrt{b^2-bc+c^2}}\le\dfrac{1}{2}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{c^2-ca+a^2}}\le\dfrac{1}{2}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\)
Cộng vế:
\(P\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(P=2(\frac{1}{ab+bc+ac}+\frac{1}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2})+\frac{1}{2(ab+bc+ac)}\\
\geq 2.\frac{9}{2(ab+bc+ac)+a^2+b^2+c^2}+\frac{1}{2(ab+bc+ac)}\\
=\frac{18}{(a+b+c)^2}+\frac{1}{2(ab+bc+ac)}\\
=18+\frac{1}{2(ab+bc+ac)}\)
Áp dụng BĐT AM-GM:
$2(ab+bc+ac)\leq 2.\frac{(a+b+c)^2}{3}=\frac{2}{3}$
$\Rightarrow \frac{1}{2(ab+bc+ac)}\geq \frac{3}{2}$
$\Rightarrow P\geq 18+\frac{3}{2}=\frac{39}{2}$
Vậậy $P_{\min}=\frac{39}{2}$ khi $a=b=c=\frac{1}{3}$
\(\dfrac{\sqrt{b^2+a^2+a^2}}{ab}\ge\dfrac{\sqrt{\dfrac{1}{3}\left(b+a+a\right)^2}}{ab}=\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{a}+\dfrac{2}{b}\right)\)
Tương tự: \(\dfrac{\sqrt{c^2+2b^2}}{bc}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{b}+\dfrac{2}{c}\right)\) ; \(\dfrac{\sqrt{a^2+2c^2}}{ac}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{c}+\dfrac{2}{a}\right)\)
Cộng vế với vế:
\(VT\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\sqrt{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1980\sqrt{3}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{3}{1980}\)