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Dự đoán điểm rơi xảy ra tại \(\left(a;b;c\right)=\left(3;2;4\right)\)
Đơn giản là kiên nhẫn tính toán và tách biểu thức:
\(D=13\left(\dfrac{a}{18}+\dfrac{c}{24}\right)+13\left(\dfrac{b}{24}+\dfrac{c}{48}\right)+\left(\dfrac{a}{9}+\dfrac{b}{6}+\dfrac{2}{ab}\right)+\left(\dfrac{a}{18}+\dfrac{c}{24}+\dfrac{2}{ac}\right)+\left(\dfrac{b}{8}+\dfrac{c}{16}+\dfrac{2}{bc}\right)+\left(\dfrac{a}{9}+\dfrac{b}{6}+\dfrac{c}{12}+\dfrac{8}{abc}\right)\)
Sau đó Cô-si cho từng ngoặc là được
Lời giải:
Ta có:
\(P=\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}=\frac{(ab)^2+(bc)^2+(ca)^2}{abc}\)
Xét tử số:
\(\text{TS}=(ab)^2+(bc)^2+(ca)^2\)
\(\Rightarrow \text{TS}^2=a^4b^4+b^4c^4+c^4a^4+2(a^2b^4c^2+a^2b^2c^4+a^4b^2c^2)\)
Áp dụng BĐT AM-GM ta có:
\(\left\{\begin{matrix} a^4b^4+b^4c^4\geq 2a^2b^4c^2\\ b^4c^4+c^4a^4\geq 2a^2b^2c^4\\ c^4a^4+a^4b^4\geq 2a^4b^2c^2\end{matrix}\right.\)
Cộng theo vế và rút gọn:
\(\Rightarrow a^4b^4+b^4c^4+c^4a^4\geq a^2b^4c^2+a^2b^2c^4+a^4b^2c^2\)
Do đó:
\(\text{TS}^2\geq 3(a^2b^4c^2+a^2b^2c^4+a^4b^2c^2)=3a^2b^2c^2(a^2+b^2+c^2)=3a^2b^2c^2\)
\(\Rightarrow \text{TS}\geq \sqrt{3}abc\)
\(\Rightarrow P\geq \sqrt{3}\)
Vậy \(P_{\min}=\sqrt{3}\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\)
Cách khác:
\(P^2=\dfrac{a^2b^2}{c^2}+\dfrac{b^2c^2}{a^2}+\dfrac{c^2a^2}{b^2}+2\left(a^2+b^2+c^2\right)\)
Áp dụng BĐT Cauchy:
\(\dfrac{a^2b^2}{c^2}+\dfrac{b^2c^2}{a^2}\ge2b^2\)
CMTT\(\Rightarrow\)\(\dfrac{a^2b^2}{c^2}+\dfrac{b^2c^2}{a^2}+\dfrac{a^2c^2}{b^2}\ge a^2+b^2+c^2\)
\(\Rightarrow P^2\ge3\Rightarrow P\ge\sqrt{3}\)
Dấu"=" xảy ra\(\Leftrightarrow\)a=b=c=\(\dfrac{1}{\sqrt{3}}\)
\(P=\sqrt{\dfrac{ab}{c+ab}}+\sqrt{\dfrac{bc}{a+bc}}+\sqrt{\dfrac{ca}{b+ca}}\)
\(=\sqrt{\dfrac{ab}{c\left(a+b+c\right)+ab}}+\sqrt{\dfrac{bc}{a\left(a+b+c\right)+bc}}+\sqrt{\dfrac{ca}{b\left(a+b+c\right)+ca}}\)
\(=\sqrt{\dfrac{ab}{\left(b+c\right)\left(c+a\right)}}+\sqrt{\dfrac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{ca}{\left(a+b\right)\left(b+c\right)}}\)
\(\le\dfrac{1}{2}\left(\dfrac{b}{b+c}+\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{a+b}+\dfrac{c}{b+c}+\dfrac{a}{a+b}\right)=\dfrac{1}{2}\)
\("=" \Leftrightarrow a=b=c=\frac{1}{3}\)
\(a+b+c=2\Rightarrow ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{4}{3}\)
\(P=\dfrac{7+2b}{1+a}+\dfrac{7+2c}{1+b}+\dfrac{7+2a}{1+c}\)
\(\ge\dfrac{\left(21+2\left(a+b+c\right)\right)^2}{\left(1+a\right)\left(7+2b\right)+\left(1+b\right)\left(7+2c\right)+\left(1+c\right)\left(7+2a\right)}\)
\(=\dfrac{25^2}{21+9\left(a+b+c\right)+2\left(ab+bc+ca\right)}\ge\dfrac{25^2}{21+9.2+\dfrac{2.4}{3}}=15\)
\("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)
Ta có : \(9=a^2+a^2+b^2+a^2+b^2+bc+bc+c^2+c^2\ge9\sqrt[9]{a^6\cdot b^6\cdot c^6}=9\sqrt[3]{a^2\cdot b^2\cdot c^2}\Rightarrow abc\le1\) Áp dụng bđt Cô-si vào các số dương : \(a^2+\dfrac{1}{b^2}+\dfrac{1}{b^2}+\dfrac{1}{b^2}\ge4\sqrt[4]{\dfrac{a^2}{b^6}}=4\sqrt{\dfrac{a}{b^3}}\Rightarrow\sqrt{a^2+\dfrac{3}{b^2}}\ge2\cdot\sqrt[4]{\dfrac{a}{b^3}}\)
CM tương tự ta được: \(\sqrt{b^2+\dfrac{3}{c^2}}\ge2\sqrt[4]{\dfrac{b}{c^3}};\sqrt{c^2+\dfrac{3}{a^2}}\ge2\sqrt[4]{\dfrac{c}{a^3}}\Rightarrow P\ge2\cdot\left(\sqrt[4]{\dfrac{a}{b^3}}+\sqrt[4]{\dfrac{b}{c^3}}+\sqrt[4]{\dfrac{c}{a^3}}\right)\ge2\cdot3\cdot\sqrt[12]{\dfrac{a}{b^3}\cdot\dfrac{b}{c^3}\cdot\dfrac{c}{a^3}}=6\sqrt[12]{\dfrac{1}{\left(abc\right)^2}}=6\) Dấu = xảy ra \(\Leftrightarrow a=b=c=1\)
Bài 1
\(VT=\dfrac{a^2}{ab^2+abc+ac^2}+\dfrac{b^2}{c^2b+abc+a^2b}+\dfrac{c^2}{a^2c+abc+b^2c}\)
Áp dụng bđt Cauchy dạng phân thức
\(\Rightarrow VT\ge\dfrac{\left(a+b+c\right)^2}{ab\left(a+b\right)+abc+ac\left(a+c\right)+abc+bc\left(b+c\right)+abc}\)
\(\Leftrightarrow VT\ge\dfrac{\left(a+b+c\right)^2}{ab\left(a+b+c\right)+ac\left(a+b+c\right)+bc\left(a+b+c\right)}=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)\left(ab+bc+ac\right)}\)
\(\Leftrightarrow VT\ge\dfrac{a+b+c}{ab+bc+ac}\left(đpcm\right)\)
Dấu ''='' xảy ra khi \(a=b=c\)
Bài 2
\(VT=\left(\sqrt{a^2}+\sqrt{b^2}+\sqrt{c^2}\right)\left[\left(\dfrac{\sqrt{a}}{b+c}\right)^2+\left(\dfrac{\sqrt{b}}{c+a}\right)^2+\left(\dfrac{\sqrt{c}}{a+b}\right)^2\right]\)
Áp dụng bđt Bunhiacopxki ta có
\(VT\ge\left(\sqrt{a}.\dfrac{\sqrt{a}}{b+c}+\sqrt{b}.\dfrac{\sqrt{b}}{c+a}+\sqrt{c}.\dfrac{\sqrt{c}}{a+b}\right)^2\)
\(\Leftrightarrow VT\ge\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2\)
Xét \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
Áp dụng bđt Cauchy dạng phân thức ta có
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}=\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ac\right)}=\dfrac{3}{2}\)
\(\Rightarrow\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2\ge\left(\dfrac{3}{2}\right)^2=\dfrac{9}{4}\)
\(\Rightarrow VT\ge\dfrac{9}{4}\left(đpcm\right)\)
Dấu '' = '' xảy ra khi \(a=b=c\)
Ta có:\(\sum\dfrac{a^2+6a+3}{a^2+a}=\sum\left(1+\dfrac{5a+3}{a^2+a}\right)=3+\sum\dfrac{5a+3}{a^2+a}\)
Có BĐT phụ: \(\dfrac{5a+3}{a^2+a}\ge-\dfrac{7}{2}a+\dfrac{15}{2}\)đúng vì nó tương đương \(\left(7a+6\right)\left(a-1\right)^2\ge0\left(true\right)\)
Áp dụng tương tự ta có:
\(VT\ge3-\dfrac{7}{2}\left(a+b+c\right)+\dfrac{15}{2}.3\ge3-\dfrac{21}{2}+\dfrac{45}{2}=15\)
Dấu = xảy ra khi a=b=c=1
Ta có:
\(\dfrac{a}{bc}+\dfrac{b}{ca}\ge2\sqrt{\dfrac{ab}{abc^2}}=\dfrac{2}{c}\)
Tương tự: \(\dfrac{a}{bc}+\dfrac{c}{ab}\ge\dfrac{2}{b}\) ; \(\dfrac{b}{ca}+\dfrac{c}{ab}\ge\dfrac{2}{a}\)
Cộng vế với vế: \(\Rightarrow\dfrac{a}{bc}+\dfrac{b}{ca}+\dfrac{c}{ab}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Rightarrow P\ge\dfrac{a^2+b^2+c^2}{2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Rightarrow P\ge\dfrac{1}{2}\left(a^2+\dfrac{1}{a}+\dfrac{1}{a}\right)+\dfrac{1}{2}\left(a^2+\dfrac{1}{b}+\dfrac{1}{b}\right)+\dfrac{1}{2}\left(c^2+\dfrac{1}{c}+\dfrac{1}{c}\right)\)
\(\Rightarrow P\ge\dfrac{1}{2}.3\sqrt[3]{\dfrac{a^2}{a^2}}+\dfrac{1}{2}.3\sqrt[3]{\dfrac{b^2}{b^2}}+\dfrac{1}{2}.3\sqrt[3]{\dfrac{c^2}{c^2}}=\dfrac{9}{2}\)
\(P_{min}=\dfrac{9}{2}\) khi \(a=b=c=1\)