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Áp dụng bất đăng thức Holder, ta có
\(\Sigma_{cyc} a \sqrt[3]{b^2+c^2} = \Sigma_{cyc} \sqrt[3]{a.a^2.(b^2+c^2)} \le \sqrt[3]{( \Sigma_{cyc} a).(\Sigma_{cyc} a^2).[\Sigma_{cyc} (b^2+c^2)} \le \sqrt[3]{\sqrt{3\Sigma_{cyc} a^2}.(\Sigma_{cyc} a^2).(2\Sigma_{cyc} a^2}) \le 12\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Điều kiện đã cho
\(\Leftrightarrow\dfrac{1}{1+a}=\left(1-\dfrac{1}{1+b}\right)+\left(1-\dfrac{1}{1+c}\right)\)
\(\Leftrightarrow\dfrac{1}{1+a}=\dfrac{b}{1+b}+\dfrac{c}{1+c}\)
\(\Leftrightarrow\dfrac{1}{1+a}=\dfrac{b+c+2bc}{bc+b+c+1}\)
\(\Leftrightarrow bc+b+c+1=b+c+2bc+ab+ac+2abc\)
\(\Leftrightarrow2abc+ab+bc+ca=1\)
Mà \(ab+bc+ca\ge3\left(\sqrt[3]{abc}\right)^2\)
\(\Rightarrow2abc+3\left(\sqrt[3]{abc}\right)^2\le1\)
Đặt \(\sqrt[3]{abc}=t\left(t\ge0\right)\), khi đó \(2t^3+3t^2\le1\)
\(\Leftrightarrow\left(t+1\right)^2\left(2t-1\right)\le0\)
Do \(\left(t+1\right)^2\ge0\) nên \(2t-1\le0\) \(\Leftrightarrow t\le\dfrac{1}{2}\) \(\Leftrightarrow abc\le\dfrac{1}{8}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{2}\)
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Vì \(-1\le a\le1\Rightarrow a^2\le1\)Tương tự có \(b^2\le1;c^2\le1\)
Suy ra \(P=a^2+2b^2+c^2\le1+2\cdot1+1=4\)hay \(maxP=4\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\pm1\)
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Câu 2a
\(\left(ac+bd\right)^2+\left(ad-bc\right)^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2=\left(a^2+b^2\right)c^2+d^2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2c^2+b^2d^2+a^2d^2+b^2c^2=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)
\(\Leftrightarrow a^2c^2+b^2d^2+a^2d^2+b^2c^2-\left(a^2c^2+b^2d^2+a^2d^2+b^2c^2\right)=0\)
\(\Leftrightarrow0=0\)( đpcm )
Câu 2b
\(\left(ac+bd\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2c^2+2abcd+b^2d^2\le\left(a^2+b^2\right)c^2+d^2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2c^2+2abcd+b^2d^2\le a^2c^2+b^2c^2+a^2d^2+b^2d^2\)
\(\Leftrightarrow2abcd\le b^2c^2+a^2d^2\)
\(\Leftrightarrow0\le b^2c^2-2abcd+a^2d^2\)
\(\Leftrightarrow0\le\left(bc-ad\right)^2\)( đpcm )
Câu 4a
\(\frac{a+b}{2}\ge\sqrt{ab}\)
\(\Leftrightarrow\left(\frac{a+b}{2}\right)^2\ge ab\)
\(\Leftrightarrow\frac{\left(a+b\right)^2}{4}\ge ab\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow a^2+2ab+b^2\ge4ab\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)( đpcm )
Câu 4c
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow3a+5b\ge2\sqrt{3a.5b}=2\sqrt{15ab}\)
\(\Rightarrow12\ge2\sqrt{15ab}\)
\(\Rightarrow6\ge\sqrt{15ab}\)
\(\Rightarrow6^2\ge15ab\)
\(\Rightarrow36\ge15ab\)
\(\Rightarrow ab\le\frac{12}{5}\)
\(\Leftrightarrow P\le\frac{12}{5}\)
Vậy GTLN của \(P=\frac{12}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 5 nhé:
a) (a+1)2>=4a
<=>a2+2a+1>=4a
<=>a2-2a+1.>=0
<=>(a-1)2>=0 (luôn đúng)
vậy......
b) áp dụng bất dẳng thức cô si cho 2 số dương 1 và a ta có:
a+1>=\(2\sqrt{a}\)
tương tự ta có:
b+1>=\(2\sqrt{b}\)
c+1>=\(2\sqrt{c}\)
nhân vế với vế ta có:
(a+1)(b+1)(c+1)>=\(2\sqrt{a}.2\sqrt{b}.2\sqrt{c}\)
<=>(a+1)(b+1)(c+1)>=\(8\sqrt{abc}\)
<=>(a+)(b+1)(c+1)>=8 (vì abc=1)
vậy....
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\(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)=9\Rightarrow-3\le a+b+c\le3\)
\(S=a+b+c+\dfrac{\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)}{2}=\dfrac{1}{2}\left(a+b+c\right)^2+a+b+c-\dfrac{3}{2}\)
Đặt \(a+b+c=x\Rightarrow-3\le x\le3\)
\(S=\dfrac{1}{2}x^2+x-\dfrac{3}{2}=\dfrac{1}{2}\left(x+1\right)^2-2\ge-2\)
\(S_{min}=-2\) khi \(\left\{{}\begin{matrix}a+b+c=-1\\a^2+b^2+c^2=3\end{matrix}\right.\) (có vô số bộ a;b;c thỏa mãn)
\(S=\dfrac{1}{2}\left(x^2+2x-15\right)+6=\dfrac{1}{2}\left(x-3\right)\left(x+5\right)+6\le6\)
\(S_{max}=6\) khi \(x=3\) hay \(a=b=c=1\)
\(a,b,c\le2\)nên \(\left(a-2\right)\left(b-2\right)\left(c-2\right)\le0\)
\(\Leftrightarrow abc-2\left(ab+bc+ca\right)+4\left(a+b+c\right)-8\le0\)
\(\Leftrightarrow ab+bc+ca\ge\frac{4.3-8-abc}{2}\ge2\)(do \(a,b,c\ge0\)nên \(abc\ge0\))
\(A=a^2+b^2+c^2=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\le3^2-2.2=5\)
Dấu \(=\)khi \(\hept{\begin{cases}a+b+c=3\\abc=0\\\left(a-2\right)\left(b-2\right)\left(c-2\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=0\\b=1\\c=2\end{cases}}\)và các hoán vị.