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\(15\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+30\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=40\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2007\)
\(\Leftrightarrow15\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=40\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2007\)
\(\Leftrightarrow15\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le\frac{40}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2+2007\)
\(\Leftrightarrow\frac{5}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le2007\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\sqrt{\frac{6021}{5}}\)
Ta có:
\(5a^2+2ab+2b^2=4a^2+2ab+b^2+a^2+b^2\ge4a^2+2ab+b^2+2ab=\left(2a+b\right)^2\)
\(\Rightarrow\sqrt{5a^2+2ab+2b^2}\ge2a+b\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)
\(\Rightarrow P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}=\frac{1}{a+a+b}+\frac{1}{b+b+c}+\frac{1}{c+c+a}\)
\(\Rightarrow P\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow P\le\frac{1}{3}\sqrt{\frac{6021}{5}}\)
Dấu "=" xảy ra khi \(a=b=c=3\sqrt{\frac{5}{6021}}\)
Mẫu thức như vầy thì tìm max còn được chứ tìm min sao nổi bạn?
+)\(\frac{3}{4}\ge a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\Leftrightarrow\frac{1}{8}\ge abc\)
+) \(P=8abc+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\left(32abc+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)-24abc\)
\(\ge4\sqrt[4]{\frac{32}{abc}}-24abc\ge4\sqrt[4]{\frac{32}{\frac{1}{8}}}-3=16-3=13\)
Dấu = xảy ra khi \(a=b=c=\frac{1}{2}\)
Câu 1:
\(\left(a+b+c\right)^2=a^2+b^2+c^2\Rightarrow ab+ac+bc=0\Rightarrow bc=-ab-ac\)
\(a^2+2bc=a^2+bc+bc=a^2+bc-ac-ab=\left(a-b\right)\left(a-c\right)\)
Tương tự: \(b^2+2ac=\left(b-a\right)\left(b-c\right)\); \(c^2+2ab=\left(a-c\right)\left(b-c\right)\)
\(P=\frac{a^2}{\left(a-b\right)\left(a-c\right)}-\frac{b^2}{\left(a-b\right)\left(b-c\right)}+\frac{c^2}{\left(a-c\right)\left(b-c\right)}=\frac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(P=\frac{a^2\left(b-c\right)-b^2a+ac^2+b^2c-bc^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\frac{a^2\left(b-c\right)-\left(ab+ac\right)\left(b-c\right)+bc\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(P=\frac{\left(b-c\right)\left(a^2-ab-ac+bc\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\frac{\left(b-c\right)\left(a-b\right)\left(a-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)
Câu 2:
\(x=a+1\); \(y=4\left(a+1\right)^2+1=4x^2+1\); \(z=6\left(a+1\right)^2+1=6x^2+1\)
- Nếu \(x=2\Rightarrow z=25\) không phải nguyên tố (loại)
- Nếu \(x=3\Rightarrow z=55\) không phải nguyên tố (loại)
- Nếu \(x=5\Rightarrow\left\{{}\begin{matrix}y=101\\z=151\end{matrix}\right.\) là số nguyên tố \(\Rightarrow a=4\)
- Nếu \(x>5\) ta có các trường hợp:
+) \(x=5k+1\Rightarrow y=4\left(5k+1\right)^2+1=4\left(25k^2+10k\right)+5⋮5\) (loại)
+) \(x=5k+2\Rightarrow z=6\left(5k+2\right)^2+1=6\left(25k^2+20k\right)+25⋮25\) (loại)
+) \(x=5k+3\Rightarrow z=6\left(25k^2+30k\right)+55⋮5\) (loại)
+) \(x=5k+4\Rightarrow y=4\left(25k^2+40k\right)+65⋮5\) (loại)
Vậy \(a=4\) là số tự nhiên duy nhất thỏa điều kiện đề bài
Lời giải:
\(a+b+c=0\Rightarrow b+c=-a\). Khi đó:
\(a^2-b^2-c^2=a^2-(b^2+c^2)=a^2-(b^2+2bc+c^2)+2bc\)
\(=a^2-(b+c)^2+2bc=a^2-(-a)^2+2bc=2bc\)
\(\Rightarrow \frac{a^2-2bc}{a^2-b^2-c^2}=\frac{a^2-2bc}{2bc}=\frac{a^2}{2bc}-2\)
Hoàn toàn tương tự với các phân thức còn lại:
\(P=\frac{a^2}{2bc}-2+\frac{b^2}{2ac}-2+\frac{c^2}{2ab}-2\)
\(=\frac{a^3+b^3+c^3}{2bac}-6=\frac{(a+b)^3-3ab(a+b)+c^3}{2abc}-6\)
\(=\frac{(-c)^3-3ab(-c)+c^3}{2abc}-6=\frac{3abc}{2abc}-6=\frac{3}{2}-6=\frac{-9}{2}\)
Làm tạm một câu rồi đi chơi, lát làm cho.
4)
Áp dụng bất đẳng thức Cauchy-Schwarz :
\(VT\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=\frac{9}{\left(a+b+c\right)^2}\ge\frac{9}{1}=9\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
\(P=\frac{1}{a^2+b^2+c^2}+\frac{a+b+c}{abc}=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
\(P\ge\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+bc+ca}=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}+\frac{7}{ab+bc+ca}\)
\(P\ge\frac{9}{a^2+b^2+c^2+ab+bc+ca+ab+bc+ca}+\frac{7}{\frac{\left(a+b+c\right)^2}{3}}=\frac{30}{\left(a+b+c\right)^2}=30\)
\(P_{min}=30\) khi \(a=b=c=\frac{1}{3}\)
Ta thấy:
\(\left(a^2+2bc\right)+\left(b^2+2ac\right)+\left(c^2+2ab\right)=\left(a+b+c\right)^2\le1\)
Áp dụng BĐT AM-GM ta có:
\(P\ge\left[\left(a^2+2bc\right)+\left(b^2+2ac\right)+\left(c^2+2ab\right)\right]\left(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\right)\)
\(\ge3\sqrt[3]{\left(a^2+2bc\right)\left(b^2+2ac\right)\left(c^2+2ab\right)}\cdot3\sqrt[3]{\frac{1}{a^2+2bc}\cdot\frac{1}{b^2+2ac}\cdot\frac{1}{c^2+2ab}}=9\)
Dấu "="xảy ra khi \(\left\{\begin{matrix}a+b+c=1\\a^2+2bc=b^2+2ac=c^2+2ab\end{matrix}\right.\)\(\Rightarrow a=b=c=\frac{1}{3}\)
Vậy \(Min_P=9\) khi \(a=b=c=\frac{1}{3}\)