tim gtnn
cho a+b+c=6
tim gtnn
A=(a+1)/a+(b+1)/b+(c+4)/c
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Áp dụng BĐT AM-GM ta có:
\(A=5a+6b+7c+\frac{1}{a}+\frac{8}{b}+\frac{27}{c}\)
\(=4\left(a+b+c\right)+\left(\frac{1}{a}+a\right)+\left(\frac{8}{b}+2b\right)+\left(\frac{27}{c}+3c\right)\)
\(\ge4\cdot6+2\sqrt{\frac{1}{a}\cdot a}+2\sqrt{\frac{8}{b}\cdot2b}+2\sqrt{\frac{27}{c}\cdot3c}\)
\(\ge24+2+2\cdot4+2\cdot9=52\)
Xảy ra khi \(\frac{1}{a}=a;\frac{8}{b}=2b;\frac{27}{c}=3c\Rightarrow a=1;b=2;c=3\)
\(A=\frac{a^2}{a+b}+\frac{b^2}{c+a}+\frac{c^2}{b+c}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}=\frac{6}{2}=3..\)
\(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\Leftrightarrow b=\frac{2ac}{a+c}\)
\(P=\frac{a+b}{2a-b}+\frac{b+c}{2c-b}=\frac{a+\frac{2ac}{a+c}}{2a-\frac{2ac}{a+c}}+\frac{\frac{2ac}{a+c}+c}{2c-\frac{2ac}{a+c}}=\frac{a+3c}{2a}+\frac{3a+c}{2c}=1+\frac{3}{2}\left(\frac{a}{c}+\frac{c}{a}\right)\ge4\)
Dấu "=" xảy ra khi \(a=b=c\)
Nhân P với 4. Do 4=a+b+c+d+e
Áp dụng \(\left(x+y\right)^2\ge4xy\)
Lời giải:
Ta có: \(A=\frac{a+1}{a}+\frac{b+1}{b}+\frac{c+4}{c}\)
\(\Leftrightarrow A=1+\frac{1}{a}+1+\frac{1}{b}+1+\frac{4}{c}=3+\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\)
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)(a+b+c)\geq (1+1+2)^2\)
\(\Leftrightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\geq \frac{4^2}{a+b+c}=\frac{16}{6}=\frac{8}{3}\)
Do đó: \(A\geq 3+\frac{8}{3}=\frac{17}{3}\) hay \(A_{\min}=\frac{17}{3}\)
Dấu bằng xảy ra khi \((a,b,c)=(\frac{3}{2}; \frac{3}{2}; 3)\)
cai hang thu ba la dung bat dang gi vay ban hoi do gioi khong thay