Tim min ,max.......

N

nguyenbahiep1

bai 2:cho [TEX]a\geq2 ,b\geq3, c\geq4[/TEX] .Tim GTLN :
[TEX]P=\sqrt{c-4}/c +\sqrt{a-2}/a + \sqrt{b-3}/b[/TEX]

[laTEX]\frac{\sqrt{c-4}}{c} \leq \frac{4+c-4}{4c} = \frac{1}{4} \\ \\ \frac{\sqrt{a-2}}{a} \leq \frac{2+a-2}{2\sqrt{2}c} = \frac{\sqrt{2}}{4} \\ \\ \frac{\sqrt{b-3}}{b} \leq \frac{3+b-3}{2\sqrt{3}c} = \frac{\sqrt{3}}{6}[/laTEX]

cộng dồn 3 vế vào là ra đáp án
 
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hoang_duythanh

S=$\frac{1}{a^2+b^2+c^2}+\frac{1}{3ab}+\frac{1}{3bc}+\frac{1}{3ac}+\frac{2}{3ab}+\frac{2}{3bc}+\frac{2}{3ac}$
\geq$\frac{16}{a^2+b^2+c^2+3ab+3bc+3ca}+\frac{2}{3}.\frac{9}{ab+bc+ca}$(cauchy-schawarz)
Có $ab+bc+ca$\leq$\frac{(a+b+c)^2}{3}$\leq$\frac{1}{3}$
=>$a^2+b^2+c^2+3ab+3bc+3ca$=$(a+b+c)^2+ab+bc+ca$ \leq $\frac{4}{3}$
=>S\geq 12+18=30
dấu "=" khi a=b=c=$\frac{1}{3}$
 
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