Для положительных $a,b,c$ докажите неравество

                   $\frac{1}{a + 2b + 3c} + \frac{1}{b + 2c + 3a} + \frac{1}{c + 2a + 3b} \leq \frac{1}{6}(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$

  Proposed by Saken Ilyassov

$a,b,c\ge 0$. Daleldeniz: \[(a+b+c+1)(a+1)(b+1)(c+1)\ge 8(ab+bc+ca+abc)\]

$\triangle ABC$  uwburiwinda, $\angle A \le 90^o$. Daleldeniz:

$$\cos (B-C)\le \frac{2bc}{b^2+c^2}$$

Negizi bul da "tema" esep (oblis 2011, 10-sinip). Men osi esepke ademi kiska wewim taptim. Aldimen sizderdin wewimderinizdi korsem, keyin ozimdikin jazamin. Oblistik olimpiada kezinde okuwilar buni wigargan wigar? Al boliseyik :)

Vietnamnin esebinen mikti:

$a,b,c>0, \ \ \ abc=1$.

Daleldeniz:

$$a^3+b^3+c^3+6\ge (a+b+c)^2+(a+b+c-3)^2$$