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\(P=\frac{1}{25a}+\frac{1}{16b}+\frac{1}{9c}=\frac{\frac{1}{25}}{a}+\frac{\frac{1}{16}}{b}+\frac{\frac{1}{9}}{c}\ge\frac{\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}\right)^2}{a+b+c}=\frac{2209}{3600}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\frac{\frac{1}{5}}{a}=\frac{\frac{1}{4}}{b}=\frac{\frac{1}{3}}{c}=\frac{\frac{1}{5}+\frac{1}{4}+\frac{1}{3}}{a+b+c}=\frac{47}{60}\)
\(\Rightarrow\)\(\hept{\begin{cases}a=\frac{1}{5}:\frac{47}{60}=\frac{12}{47}\\b=\frac{1}{4}:\frac{47}{60}=\frac{15}{47}\\c=\frac{1}{3}:\frac{47}{60}=\frac{20}{47}\end{cases}}\)
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BĐT phụ:\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Leftrightarrow\left(x-y\right)^2\ge0\left(true\right)\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{4}{a+b}+\frac{1}{c}\ge\frac{9}{a+b+c}\) ( đpcm )
Vậy.......
Vì \(a,b,c>0\)\(\Rightarrow\frac{a}{b};\frac{b}{c};\frac{c}{a}>0\)nên áp dụng bđt Cauchy cho 3 số dương ta có
\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3.\sqrt[3]{\frac{a}{b}.\frac{b}{c}.\frac{c}{a}}=3.\sqrt[3]{1}=3\left(đpcm\right)\)
Vậy với \(a,b,c>0\)thì \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\)
Áp dụng BĐT AM-GM ta có:
\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3.\sqrt[3]{\frac{a}{b}.\frac{b}{c}.\frac{c}{a}}=3.\sqrt[3]{1}=3\)
đpcm
Xét hiệu:
\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-\frac{c}{b}-\frac{a}{c}=\frac{a-c}{b}+\frac{b-a}{c}+\frac{c-b}{a}\)
\(=\frac{ca.\left(a-c\right)}{abc}+\frac{ab.\left(b-a\right)}{abc}+\frac{bc.\left(c-b\right)}{abc}\)\(=\frac{a^2c-c^2a}{abc}+\frac{b^2a-a^2b}{abc}+\frac{c^2b-b^2c}{abc}\)
\(=\frac{a^2c-c^2a+b^2a-a^2b+c^2b-b^2c}{abc}\)\(=\frac{\left(a^2c-b^2c\right)+\left(-c^2a+c^2b\right)+\left(b^2a-a^2b\right)}{abc}\)
\(=\frac{c.\left(a-b\right)\left(a+b\right)-c^2.\left(a-b\right)-ab.\left(a-b\right)}{abc}\)\(=\frac{\left(a-b\right)\left[c.\left(a+b\right)-c^2-ab\right]}{abc}\)
\(=\frac{\left(a-b\right)\left(ac+bc-c^2-ab\right)}{abc}\)\(=\frac{\left(a-b\right)\left[\left(ac-c^2\right)+\left(bc-ab\right)\right]}{abc}\)
\(=\frac{\left(a-b\right)\left[c.\left(a-c\right)-b.\left(a-c\right)\right]}{abc}\)\(=\frac{\left(a-b\right)\left(a-c\right)\left(c-b\right)}{abc}\)
ta thấy \(a\ge b\ge c>0\Rightarrow abc>0\)
\(a-b\ge0\left(a\ge b\right);a-c\ge0\left(a\ge b\ge c\right);c-b\le0\left(b\ge c\right)\)\(\Rightarrow\left(a-b\right)\left(a-c\right)\left(c-b\right)\le0\)
\(\text{Suy ra: }\frac{\left(a-b\right)\left(a-c\right)\left(c-b\right)}{abc}\le0\)
\(\Rightarrow\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\le\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\)
có thể sai đề
\(\sqrt{\frac{a}{c+b}}=\frac{a}{\sqrt{a\left(c+b\right)}}\ge\frac{a}{\frac{a+b+c}{2}}=\frac{2a}{a+b+c}\)
tương tự : \(\sqrt{\frac{b}{a+c}}\ge\frac{2b}{a+b+c};\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\)
\(\Rightarrow\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge\frac{2\left(a+b+c\right)}{a+b+c}=2\)(ĐPCM)
\(VT=\frac{25a}{b+c}+25+\frac{16b}{c+a}+16+\frac{c}{a+b}+1-42\)
\(VT=\frac{25\left(a+b+c\right)}{b+c}+\frac{16\left(a+b+c\right)}{c+a}+\frac{a+b+c}{a+b}-42\)
\(VT=\left(a+b+c\right)\left(\frac{25}{b+c}+\frac{16}{c+a}+\frac{1}{a+b}\right)-42\)
\(VT\ge\left(a+b+c\right).\frac{\left(5+4+1\right)^2}{2\left(a+b+c\right)}-42=8\)
Bạn dùng BĐT gì vậy ạ?