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\(a\ge2b\Rightarrow\dfrac{a}{b}\ge2\)
\(A=\dfrac{a}{b}+\dfrac{b}{a}=\dfrac{a}{4b}+\dfrac{b}{a}+\dfrac{3}{4}.\dfrac{a}{b}\ge2\sqrt{\dfrac{ab}{4ab}}+\dfrac{3}{4}.2=\dfrac{5}{2}\)
\(A_{min}=\dfrac{5}{2}\) khi \(a=2b\)
Hi vọng là tìm GTLN:
Không mất tính tổng quát, giả sử b, c cùng phía với 1 \(\Rightarrow\left(b-1\right)\left(c-1\right)\ge0\Leftrightarrow bc\ge b+c-1\).
Áp dụng bất đẳng thức AM - GM ta có:
\(4=a^2+b^2+c^2+abc\ge a^2+2bc+abc\Leftrightarrow2bc+abc\le4-a^2\Leftrightarrow bc\left(a+2\right)\le\left(2-a\right)\left(a+2\right)\Leftrightarrow bc+a\le2\)
\(\Rightarrow a+b+c\le3\).
Áp dụng bất đẳng thức Schwarz ta có:
\(P\le\dfrac{ab}{9}\left(\dfrac{1}{a}+\dfrac{2}{b}\right)+\dfrac{bc}{9}\left(\dfrac{1}{b}+\dfrac{2}{c}\right)+\dfrac{ca}{9}\left(\dfrac{1}{c}+\dfrac{2}{a}\right)=\dfrac{1}{9}.3\left(a+b+c\right)=\dfrac{1}{3}\left(a+b+c\right)\le1\).
Đẳng thức xảy ra khi a = b = c = 1.
Theo đề ra, ta có:
\(a^2+b^2+c^2\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)
\(=a^3+b^3+c^3+a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\)
Theo BĐT Cô-si:
\(\left\{{}\begin{matrix}a^3+ab^2\ge2a^2b\\b^3+bc^2\ge2b^2c\\c^3+ca^2\ge2c^2a\end{matrix}\right.\Rightarrow a^2+b^2+c^2\ge3\left(a^2b+b^2c+c^2a\right)\)
Do vậy \(M\ge14\left(a^2+b^2+c^2\right)+\dfrac{3\left(ab+bc+ac\right)}{a^2+b^2+c^2}\)
Ta đặt \(a^2+b^2+c^2=k\)
Luôn có \(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2=1\)
Vì thế nên \(k\ge\dfrac{1}{3}\)
Khi đấy:
\(M\ge14k+\dfrac{3\left(1-k\right)}{2k}=\dfrac{k}{2}+\dfrac{27k}{2}+\dfrac{3}{2k}-\dfrac{3}{2}\ge\dfrac{1}{3}.\dfrac{1}{2}+2\sqrt{\dfrac{27k}{2}.\dfrac{3}{2k}}-\dfrac{3}{2}=\dfrac{23}{3}\)
\(\Rightarrow Min_M=\dfrac{23}{3}\Leftrightarrow a=b=c=\dfrac{1}{3}\).
\(\left(a^3+b\right)\left(\dfrac{1}{a}+b\right)\ge\left(a+b\right)^2\Rightarrow\dfrac{1}{a^3+b}\le\dfrac{\dfrac{1}{a}+b}{\left(a+b\right)^2}=\dfrac{ab+1}{a\left(a+b\right)^2}\)
Tương tự: \(\dfrac{1}{b^3+a}\le\dfrac{ab+1}{b\left(a+b\right)^2}\)
\(\Rightarrow P\le\left(a+b\right)\left(\dfrac{ab+1}{a\left(a+b\right)^2}+\dfrac{ab+1}{b\left(a+b\right)^2}\right)-\dfrac{1}{ab}\)
\(P\le\dfrac{\left(ab+1\right)}{a+b}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{1}{ab}=\dfrac{ab+1}{ab}-\dfrac{1}{ab}=1\)
\(P_{max}=1\) khi \(a=b=1\)
Áp dụng bất đẳng thức Cô - si ta có:
\(S\) \(=\) \(ab+\dfrac{1}{ab}\ge2\sqrt{ab.\dfrac{1}{ab}}\)
\(S\) \(=\) \(ab+\dfrac{1}{ab}\ge2\sqrt{1}=2\)
Dấu " = " xảy ra khi \(\left\{{}\begin{matrix}ab=\dfrac{1}{ab}\\a+b=1\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}\left(ab\right)^2=1\\a+b=1\end{matrix}\right.\)
⇔ \(a=b=0,5\)
GTNN của \(S=ab+\dfrac{1}{ab}=2\) khi \(a=b=0,5\)
S=\(ab+\dfrac{1}{ab}\)
Ta có :
Áp dụng BĐT Cauchy(cô-sy),ta có
1\(\ge a+b\ge2\sqrt{ab}\)\(\Leftrightarrow\sqrt{ab}\le\dfrac{1}{2}\)\(\Rightarrow ab\le\dfrac{1}{4}\)
Đặt x=ab(x\(\le\dfrac{1}{4}\))
\(\Rightarrow x+\dfrac{1}{x}=x+\dfrac{1}{16x}+\dfrac{15}{16x}\)
Áp dụng BĐT Cauchy (Cô -si):
\(S\ge2\sqrt{\dfrac{1}{16}}+\dfrac{15}{16x}=\dfrac{1}{2}+\dfrac{15}{16X}\ge\dfrac{1}{2}+\dfrac{16}{16.\dfrac{1}{4}}=\dfrac{17}{4}\)
Vậy Min S=\(\dfrac{17}{4}\) \(\Leftrightarrow\left\{{}\begin{matrix}a+b=1\\ab=\dfrac{1}{16ab}\\ab=\dfrac{1}{4}\\\end{matrix}\right.\) \(\Leftrightarrow a=b=\dfrac{1}{2}\)
