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Lời giải:
Áp dụng BĐT Cô-si:
$a^2+b^2\geq 2\sqrt{a^2b^2}=2|ab|\geq 2ab$
$b^2+c^2\geq 2bc$
$c^2+a^2\geq 2ac$
Cộng theo vế các BĐT trên ta được:
$2(a^2+b^2+c^2)\geq 2(ab+bc+ac)$
$\Rightarrow ab+bc+ac\leq a^2+b^2+c^2=27$
Vậy GTLN của $P$ là $27$
1a
\(A=\frac{3}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^4+b^4}{2}\ge\frac{6}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^2+b^2\right)^2}{2}}{2}\)
\(\ge10+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{4}=10+\frac{1}{16}=\frac{161}{16}\)
Dau '=' xay ra khi \(a=b=\frac{1}{2}\)
Vay \(A_{min}=\frac{161}{16}\)
1b.\(B=\frac{1}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^8+b^8}{4}\ge\frac{2}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^4+b^4\right)^2}{2}}{4}\)
\(\ge6+\frac{\left[\frac{\left(a^2+b^2\right)^2}{2}\right]^2}{8}\ge6+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{32}=6+\frac{1}{128}=\frac{769}{128}\)
Dau '=' xay ra khi \(a=b=\frac{1}{2}\)
Vay \(B_{min}=\frac{769}{128}\)khi \(a=b=\frac{1}{2}\)
Ta có: \(\sqrt{2a+bc}=\sqrt{a^2+ab+ac+bc}=\sqrt{\left(a+b\right)\left(a+c\right)}\le\frac{a+b+a+c}{2}\)
C/m tương tự \(\sqrt{2b+ac}\le\frac{b+a+b+c}{2}\)
\(\sqrt{2c+ab}\le\frac{c+a+c+b}{2}\)
\(\Rightarrow Q\le\frac{a+b+a+c+b+a+b+c+c+a+c+b}{2}=\frac{4\left(a+b+c\right)}{2}=4\)
Dấu "=" khi a = b = c = 2/3
Áp dụng bđt cô si ta có:
\(a^2+2b^2+3=\left(a^2+b^2\right)+\left(b^2+1\right)+2\ge2ab+2b+2=2\left(ab+b+1\right)\)
\(b^2+2c^2+3\ge2\left(bc+c+1\right)\)
\(c^2+2a^2+3\ge2\left(ac+a+1\right)\)
=> \(M\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)\)
\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{bcab+abc+ab}+\frac{b}{abc+ab+b}\right)\)
\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{b+1+ab}+\frac{b}{1+ab+b}\right)\)
\(=\frac{1}{2}.\frac{ab+b+1}{ab+b+1}=\frac{1}{2}\)
Bổ sung:
Dấu "=" xảy ra <=> a = b = c = 1
Vậy GTLN của M = 1/2 tại a = b = c = 1.
\(7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+3\ge7\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\le3\)Áp dụng BĐT AM-GM ta có :
\(A=\frac{1}{\sqrt{a^3+b^3+1}}+\frac{1}{\sqrt{b^3c^3+1+1}}+\frac{4\sqrt{3}}{c^6+1+2a^3+8}\)
\(\le\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{4\sqrt{3}}{2c^3+2a^3+8}=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{c^3+a^3+4}\)
\(=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{c^3+a^3+1+1+1+1}\)
\(\le\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{6\sqrt{ac}}=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{1}{\sqrt{3ac}}\)\(=\frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ac}}+\frac{1}{\sqrt{bc}}\right)\)
\(\le\frac{1}{\sqrt{3}}\sqrt{3\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)}=\sqrt{\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)}\le\sqrt{3}\) (Bunhiacopxki)
Dấu "=" xảy ra\(\Leftrightarrow a=b=c=1\)
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\(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right).\)(áp dụng bất đẳng thức bunhiacopxki)
\(\Leftrightarrow\left(a+b+c\right)^2\le3.64\Rightarrow\left(a+b+c\right)\le8\sqrt{3}\)
Lại có \(\left(ab+bc+ac\right)^2\le\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)\)(bất đẳng thức bunhiacopxki)
\(\Leftrightarrow ab+bc+ac\le a^2+b^2+c^2=64\)
Khi đó \(P=ab+bc+ca+a+b+c\le64+8\sqrt{3}\)
Dấu = xảy ra khi \(\hept{\begin{cases}a=b=c\\a^2+b^2+c^2=64\end{cases}\Leftrightarrow}a=b=c=\frac{8\sqrt{3}}{3}\)
\(M=4.\dfrac{a}{2}.\dfrac{b\sqrt{3}}{2}+a^2\le2\left(\dfrac{a^2}{4}+\dfrac{3b^2}{4}\right)+a^2=\dfrac{3}{2}\left(a^2+b^2\right)=\dfrac{3}{2}\)
\(M_{max}=\dfrac{3}{2}\) khi \(\left(a;b\right)=\left(\dfrac{\sqrt{3}}{2};\dfrac{1}{2}\right);\left(-\dfrac{\sqrt{3}}{2};-\dfrac{1}{2}\right)\)