Cho a,b,c > 1. Tìm GTNN của biểu thức: \(P=\frac{a^2}{a-1}+\frac{2b^2}{b-1}+\frac{3c^2}{c-1}\)
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23 tháng 11 2016
Ta có
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}.\frac{3}{\sqrt[3]{abc}}\ge9\)
Dấu = xảy ra khi \(a=b=c=\frac{2014}{6}=\frac{1007}{3}\)
3 tháng 1 2020
P = 4a + 7b + 10c + \(\frac{4}{a}+\frac{1}{4b}+\frac{1}{9c}\)
P = \(3\left(a+2b+3c\right)+\left(a+\frac{4}{a}\right)+\left(b+\frac{1}{4b}\right)+\left(c+\frac{1}{9c}\right)\)
\(\ge3.4+2\sqrt{a.\frac{4}{a}}+2\sqrt{b.\frac{1}{4b}}+2\sqrt{c.\frac{1}{9c}}=\frac{53}{3}\)
Vây GTNN của P là \(\frac{53}{3}\)khi \(a=1;b=\frac{1}{2};c=\frac{1}{3}\)
KN
28 tháng 5 2020
Ta có: \(\frac{1+3a}{1+b^2}=\left(1+3a\right).\frac{1}{1+b^2}=\left(1+3a\right)\left(1-\frac{b^2}{1+b^2}\right)\)
\(\ge\left(1+3a\right)\left(1-\frac{b^2}{2b}\right)=\left(1+3a\right)\left(1-\frac{b}{2}\right)\)
\(=3a+1-\frac{b}{2}-\frac{3ab}{2}\)(1)
Tương tự ta có: \(\frac{1+3b}{1+c^2}=3b+1-\frac{c}{2}-\frac{3bc}{2}\)(2); \(\frac{1+3c}{1+a^2}=3c+1-\frac{a}{2}-\frac{3ca}{2}\)(3)
Cộng theo vế của 3 BĐT (1), (2), (3), ta được: \(\frac{1+3a}{1+b^2}+\frac{1+3b}{1+c^2}+\frac{1+3c}{1+a^2}\)\(\ge3\left(a+b+c\right)-\frac{a+b+c}{2}-\frac{3\left(ab+bc+ca\right)}{2}+3\)
\(=\frac{5\left(a+b+c\right)}{2}-\frac{3\left(ab+bc+ca\right)}{2}+3\)
\(\ge\frac{5.\sqrt{3\left(ab+bc+ca\right)}}{2}-\frac{3.3}{2}+3=\frac{15}{2}-\frac{9}{2}+3=6\)
Đẳng thức xảy ra khi a = b = c = 1
TN
23 tháng 8 2016
\(VT=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3c}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+a^2}\)
Ta tách VT=A+B và xét
\(A=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3c}{1+a^2}=\text{∑}\left(3a-\frac{3ab^2}{1+b^2}\right)\ge\text{∑}\left(3a-\frac{3ab}{2}\right)\)
\(B=\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+a^2}=\text{∑}\left(1-\frac{b^2}{1+b^2}\right)\ge\text{∑}\left(1-\frac{b}{2}\right)\)
\(\Rightarrow VT=A+B=3+\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\text{∑}ab=\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\ge\frac{15}{2}-\frac{3}{2}=6\)
(Do \(a+b+c\ge\sqrt{3\left(ab+bc+ca\right)}=3\))
Dấu = khi a=b=c=1
23 tháng 8 2016
\(VT=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3c}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+a^2}\)
Ta tách VT = A + b và xét :
\(A=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3c}{1+a^2}=\Sigma\left(3a-\frac{3ab^2}{1+b^2}\right)\ge\Sigma\left(3a-\frac{3ab}{2}\right)\)\(B=\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+a^2}=\Sigma\left(1-\frac{b^2}{1+b^2}\right)\ge\Sigma\left(1-\frac{b}{2}\right)\)
\(\Rightarrow VT=A+B=3+\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\Sigma ab=\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\ge\frac{15}{2}-\frac{3}{2}=6\)( Do \(a+b+c\ge\sqrt{3\left(ab+bc+ca\right)=3}\))
Dấu = khi a = b = c = 1 .
30 tháng 4 2020
\(a^2b^2c^2+\left(a+1\right)\left(1+b\right)\left(1+c\right)\ge a+b+c+ab+bc+ca+3\)
\(\Leftrightarrow\left(abc\right)^2+abc-2\ge0\Leftrightarrow\left(abc+2\right)\left(abc-1\right)\ge0\Leftrightarrow abc\ge1\)
Áp dụng BĐT Cosi ta có:
\(\frac{a^3}{\left(b+2c\right)\left(2c+3a\right)}+\frac{b+2c}{45}+\frac{2c+3a}{75}\ge3\sqrt[3]{\frac{a^3}{\left(b+2c\right)\left(2c+3b\right)}\cdot\frac{b+2c}{45}\cdot\frac{2c+3a}{75}}=\frac{a}{5}\left(1\right)\)
Tương tự ta có: \(\hept{\begin{cases}\frac{b^3}{\left(c+2a\right)\left(2a+3b\right)}+\frac{c+2a}{45}+\frac{2a+3b}{75}\ge\frac{b}{5}\left(2\right)\\\frac{c^3}{\left(a+2b\right)\left(2b+3c\right)}+\frac{a+2b}{45}+\frac{2b+3c}{75}\ge\frac{c}{5}\left(3\right)\end{cases}}\)
Từ (1)(2)(3) ta có:
\(P+\frac{2\left(a+b+c\right)}{15}\ge\frac{a+b+c}{5}\Leftrightarrow P\ge\frac{1}{15}\left(a+b+c\right)\)
Mà \(a+b+c\ge3\sqrt[3]{abc}\Rightarrow S\ge\frac{1}{5}\)
Dấu "=" xảy ra <=> a=b=c=1