cho a,b,c>0 và abc=1/(a+b+c). Tìm min P=(a+b)(a+c)
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\(\dfrac{4}{3}\le a+\sqrt{ab}+\sqrt[3]{abc}=a+\sqrt[]{\dfrac{a}{2}.2b}+\sqrt[3]{\dfrac{a}{4}.b.4c}\)
\(\le a+\dfrac{1}{2}\left(\dfrac{a}{2}+2b\right)+\dfrac{1}{3}\left(\dfrac{a}{4}+b+4c\right)=\dfrac{4}{3}\left(a+b+c\right)\)
\(\Rightarrow Q\ge1\)
\(Q_{min}=1\) khi \(\left(a;b;c\right)=\left(\dfrac{16}{21};\dfrac{4}{21};\dfrac{1}{21}\right)\)
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\(a^{2012}+b^{2012}+c^{2012}\ge3\sqrt[3]{\left(abc\right)^{2012}}=3\)
\(\Rightarrow\dfrac{1}{a^{2012}+b^{2012}+c^{2012}}\le\dfrac{1}{3}\)
\(\Rightarrow-\dfrac{1}{a^{2012}+b^{2012}+c^{2012}}\ge-\dfrac{1}{3}\)
Lại có:
\(a^{2013}+a^{2013}+...+a^{2013}\left(\text{2012 số hạng}\right)+1\ge2013\sqrt[2013]{\left(a^{2013}\right)^{2012}}=2013.a^{2012}\)
\(\Rightarrow2012.a^{2013}+1\ge2013.a^{2012}\)
Tương tự: \(2012.b^{2013}+1\ge2013.b^{2012}\) ; \(2012.c^{2013}+1\ge2013.c^{2012}\)
Cộng vế với vế:
\(\Rightarrow a^{2013}+b^{2013}+c^{2013}\ge\dfrac{2013\left(a^{2012}+b^{2012}+c^{2012}\right)-3}{2012}\)
\(\Rightarrow A\ge\dfrac{2013\left(a^{2012}+b^{2012}+c^{2012}\right)-3}{2012\left(a^{2012}+b^{2012}+c^{2012}\right)}=\dfrac{2013}{2012}-\dfrac{3}{2012}.\dfrac{1}{a^{2012}+b^{2012}+c^{2012}}\ge\dfrac{2013}{2012}-\dfrac{3}{2012}.\dfrac{1}{3}=1\)
\(A_{min}=1\) khi \(a=b=c=1\)
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Tử là mũ 2 thật hả bạn. Mũ 3 thì giải được còn mũ 2 thì vẫn chưa nghĩ ra
Ta có :
\(abc=\frac{1}{a+b+c}\)
\(\Rightarrow abc.\left(a+b+c\right)=1\)
Lai có : \(P=\left(a+b\right)\left(a+c\right)\)
\(=a^2+ab+bc+ac\)
\(=a.\left(a+b+c\right)+bc\)
Áp dụng BĐT AM - GM ta có :
P= \(a\left(a+b+c\right)+bc\ge2\sqrt{a.\left(a+b+c\right).bc}=2\sqrt{1}=2\)
Dấu " = " xảy ra \(\Leftrightarrow a.\left(a+b+c\right)=bc\)