Cho $a,b,c>0$. Chứng minh rằng $$\frac{a^3+b^3+c^3+3abc}{(a+b)(b+c)(c+a)}+\frac{(a+b+c)^2}{6(a^2+b^2+c^2)}\ge \frac{5}{4}$$
$\frac{a^3+b^3+c^3+3abc}{(a+b)(b+c)(c+a)}+\frac{(a+b+c)^2}{6(a^2+b^2+c^2)}\ge \frac{5}{4}$
#1
Đã gửi 23-05-2013 - 11:56
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#2
Đã gửi 23-05-2013 - 20:04
Mình xin trích nguyên văn lời giải của Michael Rozenberg đăng ở bên Mathlinks :
Let $a,b,c>0$. Prove that
\[\frac{a^3+b^3+c^3+3abc}{(a+b)(b+c)(c+a)}+\frac{(a+b+c)^2}{6(a^2+b^2+c^2)}\ge \frac{5}{4}\]
$\frac{a^3+b^3+c^3+3abc}{(a+b)(b+c)(c+a)}+\frac{(a+b+c)^2}{6(a^2+b^2+c^2)}\ge \frac{5}{4}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{a^3+abc}{(a+b)(a+c)(b+c)}-\frac{3}{4}\geq\frac{1}{2}-\frac{(a+b+c)^2}{6(a^2+b^2+c^2)}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{4a^3-3a^2b-3a^2c+2abc}{(a+b)(a+c)(b+c)}\geq\sum_{cyc}\frac{(a-b)^2}{6(a^2+b^2+c^2)}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{4a^3-2a^2b-2a^2c-a^2b-a^2c+2abc}{(a+b)(a+c)(b+c)}\geq\sum_{cyc}\frac{(a-b)^2}{6(a^2+b^2+c^2)}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(a-b)^2S_c$, where $S_c=\frac{2a+2b-c}{(a+b)(a+c)(b+c)}-\frac{1}{6(a^2+b^2+c^2)}$.
Let $a\geq b\geq c$.
Hence, $S_c\geq\frac{1}{(b+c)(a+c)}-\frac{1}{6(a^2+b^2+c^2)}\geq0$,
$S_b=\frac{2a+2c-b}{(a+b)(a+c)(b+c)}-\frac{1}{6(a^2+b^2+c^2)}\geq$
$\geq\frac{1}{(a+b)(b+c)}-\frac{1}{6(a^2+b^2+c^2)}\geq0$ and $(a-c)^2\geq(b-c)^2$.
Hence, $\sum_{cyc}S_c(a-b)^2\geq S_b(a-c)^2+S_a(b-c)^2\geq(b-c)^2(S_a+S_b)=$
$=(b-c)^2\left(\frac{2a+2c-b+2b+2c-a}{(a+b)(a+c)(b+c)}-\frac{1}{3(a^2+b^2+c^2)}\right)\geq$
$\geq(b-c)^2\left(\frac{a+b+c}{(a+b)(a+c)(b+c)}-\frac{1}{3(a^2+b^2+c^2)}\right)\geq0$. Done!
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