cho abc>0 thoả mãn 1/a+1/b+1/c=4.tìm giá trị lớn nhất biểu thức.m= 1/2a+b+c + 1/a+2b+c+1/a+b+2c giúp em với em cần gấp

Áp dụng BĐT $\dfrac{1}{x+y}=<\dfrac{1}{4}(\dfrac{1}{x}+\dfrac{1}{y})(x,y>0)$ ta được:

$\dfrac{1}{2a+b+c}=<\dfrac{1}{4}(\dfrac{1}{2a}+\dfrac{1}{b+c})$

Áp dụng tiếp BĐT $\dfrac{1}{x+y}=<\dfrac{1}{4}(\dfrac{1}{x}+\dfrac{1}{y})(x,y>0)$ ta được:

$\dfrac{1}{b+c}=<\dfrac{1}{4}(\dfrac{1}{b}+\dfrac{1}{c})$

$=>\dfrac{1}{2a+b+c}=<\dfrac{1}{4}(\dfrac{1}{2a}+\dfrac{1}{4b}+\dfrac{1}{4c})=\dfrac{1}{8}(\dfrac{1}{a}+\dfrac{1}{2b}+\dfrac{1}{2c})$

Tương tự:

$\dfrac{1}{a+2b+c}=<\dfrac{1}{8}(\dfrac{1}{b}+\dfrac{1}{2a}+\dfrac{1}{2c})$

$\dfrac{1}{a+b+2c}=<\dfrac{1}{8}(\dfrac{1}{c}+\dfrac{1}{2a}+\dfrac{1}{2b})$

$=>M=<\dfrac{1}{8}(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c})$

$=<\dfrac{1}{8}.2.4$

$=<1$

Dấu "$=$" xảy ra khi: $a=b=c,\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=4$

$<=> a=b=c=\dfrac{3}{4}$

$M_{max}=1<=>a=b=c=\dfrac{3}{4}$