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`Answer:`
\(\frac{a+b}{3}=\frac{b+c}{3}=\frac{c+a}{10}\)
\(\Rightarrow\frac{a+b}{3}=\frac{b+c}{3}\)
\(\Rightarrow a+b=b+c\)
\(\Rightarrow a=c\)
Mặt khác ta có: \(\frac{b+c}{3}=\frac{c+a}{10}\)
\(\Rightarrow\frac{b+c}{3}=\frac{c+c}{10}\)
\(\Rightarrow\frac{b+c}{3}=\frac{2c}{10}\)
\(\Rightarrow\frac{b+c}{3}=\frac{c}{5}\)
\(\Rightarrow5\left(b+c\right)=3c\)
\(\Rightarrow5b+5c=3c\)
\(\Rightarrow5b=-2c\)
\(\Rightarrow b=-\frac{2}{5}c\)
Có `M=11a+20b-4c+2020`
`=>M=11c+20(-2/5c)-4c+2020`
`=>M=11c-8c-4c+2020`
`=>M=-c+2020`
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áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{a+b}{3}=\frac{b+c}{5}=\frac{c+a}{10}=\frac{a+b-b-c-c-a}{-12}=\frac{c}{6}\)
\(\Rightarrow\frac{a+b}{3}=\frac{c}{6}\Rightarrow\left(a+b\right).6=3c\Rightarrow6a+6b=3c\Rightarrow3a+3b=c\Rightarrow a+b=\frac{c}{3}\)
\(\frac{b+c}{5}=\frac{c}{6}\Rightarrow6b+6c=5c\Rightarrow6b=-c\Rightarrow b=\frac{-c}{6}\)
\(\frac{c+a}{10}=\frac{c}{6}\Rightarrow6c+6a=10c\Rightarrow6a=4c\Rightarrow3a=2c\Rightarrow a=\frac{2c}{3}\)
thay vào M ta có:
\(\frac{22c}{3}+\frac{-20c}{6}-c+2017=4c-c+2017=3c+2017\)
p/s: ko chắc :))
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ta có \(\frac{11b^3-a^3}{ab+4b^2}+\frac{11c^3-b^3}{bc+4c^2}+\frac{11a^3-c^3}{ca+4a^2}=\frac{11-\left(\frac{a}{b}\right)^3}{\frac{a}{b}+4}\cdot b+\frac{11-\left(\frac{b}{c}\right)^3}{\frac{b}{c}+4}\cdot c+\frac{11-\left(\frac{c}{a}\right)^3}{\frac{c}{a}+4}\cdot a\)
khi a=b=c=1 ta thấy đẳng thức xảy ra
xét \(f\left(x\right)=\frac{11-x^3}{x+4}\)ta có \(\frac{11-x^3}{x+4}\le-x+3\Leftrightarrow\left(x-1\right)^2\left(x+1\right)\ge0\forall x>0\)
thay x bởi a/b ta được \(\frac{11-\left(\frac{a}{b}\right)^3}{\frac{a}{b}+4}\le-\frac{a}{b}+3\Leftrightarrow\frac{11b^3-a^3}{ab+4b^2}\le-a+3b\)
tương tự \(\hept{\begin{cases}\frac{11c^3-b^3}{bc+4c^2}\le-b+3c\\\frac{11ba^3-c^3}{ac+4a^2}\le-c+3a\end{cases}}\)
cộng các bđt cùng chiều ta được
\(\frac{11b^3-a^3}{ab+4b^2}+\frac{11c^3-b^3}{bc+4c^2}+\frac{11a^3-c^3}{ac+4a^2}\le2\left(a+b+c\right)=6\)
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