cho tớ tham gia vs
áp dụng bất đẳng thức AM-GM suy rộng : [TEX]a_1x_1+a_2x_2...+a_nx_n\geq(a_1+a_2..+a_n){({{x}_{1}}^{a_1}.{{x}_{2}}^{a_2}...{{x}_{n}}^{a_n})}^ {\frac{1}{a+b+c}}[/TEX]
[TEX]3.\frac{x}{3}+\frac{5}{2}.\frac{2y}{5}+2\frac{z}{2}\geq (3+\frac{5}{2}+2){(({\frac{x}{3})}^3.{(\frac{2y}{5})}^{\frac{5}{2}}).{(\frac{z}{2})}^{2})}^{\frac{2}{15}}[/TEX]
[TEX]6.\frac{x}{6}+10.\frac{2y}{5}+14\frac{z}{2}\geq(6+10+14)({({\frac{x}{3})}^{6}.{(\frac{2y}{5})}^{10}.{(\frac{z}{2})}^ {14} )}^{\frac{1}{30}}
[/TEX]
từ đó suy ra :
[TEX](x+y+z)^2(2x+4y+7z)\geq(\frac{15}{2})^2.30.\frac{x}{3}.\frac{2y}{5}.\frac{z}{2}[/TEX]
[TEX]=> (x+y+z)^2\geq{\frac{15}{2}}^2
=> x+y+z\geq\frac{15}{2}[/TEX]
hix
sao tex hư hết z
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