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|f\prime (x)|\le 1\Rightarrow -1\le f\prime (x)\le 1 -1\le f\prime(x)\Rightarrow 0\le (f(x)+x)\prime\Rightarrow f(x)+x e functie crescatoare, deci 0\le f(x)+x\le 1\Rightarrow -x\le f(x)\le 1-x (1) f\prime(x)\le 1\Rightarrow (f(x)-x)\prime\le 0\Rightarrow f(x)-x e o functie descrescatoare, deci -1\l...

k=n\Rightarrow f(x)=(f(x),n)\Rightarrow f(x)|n k=f(x)\Rightarrow f(x^{f(x)})=1\Rightarrow x^{f(x)}=e \Rightarrow ord(x)|f(x) (1) k=ord(x)\Rightarrow f(x)=(f(x),ord(x))\Rightarrow f(x)|ord(x) (2) Din relatiile (1) si (2) se deduce ca f(x)=ord(x) e singura functie care verifica cerintele problemei.

$E(x)=\frac{1}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}$$=\frac{1}{(a^2-b^2)(a^2-c^2)(x^2+a^2)}+\frac{1}{(b^2-a^2)(b^2-c^2)(x^2+b^2)}+\frac{1}{(c^2-a^2)(c^2-b^2)(x^2+c^2)}$. De aici problema iese usor.

Stie cineva cand va fi concursul?

e) $\sqrt{a-1}=\sqrt{a}\cdot\sqrt{1-\frac{1}{a}}$
$(\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1})^2=$$(\sqrt{a(1-\frac{1}{a})}+\sqrt{b(1-\frac{1}{b})}+\sqrt{c(1-\frac{1}{c})})^2$$\le(a+b+c)(3-\frac{1}{a}-\frac{1}{b}-\frac{1}{c})=a+b+c$. Prin urmare $\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\le\sqrt{a+b+c}$

d) $(a\sqrt{a^2+c^2}+c\sqrt{b^2+c^2})^2\le (a^2+(\sqrt{b^2+c^2})^2)((\sqrt{a^2+c^2})^2+c^2)$, deci $(a\sqrt{a^2+c^2}+c\sqrt{b^2+c^2})^2\le (a^2+b^2+c^2)^2$, adica $a\sqrt{a^2+c^2}+c\sqrt{b^2+c^2}\le a^2+b^2+c^2$

c) $ab+\sqrt{(1-a^2)(1-b^2)}\le ab+\frac{1-a^2+1-b^2}{2}=1-\frac{(a-b)^2}{2}\le 1$

b)$(1+4+9)(a^2+b^2+c^2)\ge (a+2b+3c)^2$$\Rightarrow {a^2+b^2+c^2}\ge\frac{(a+2b+3c)^2}{14}$ (1)
$\frac{(a+2b+3c)^2}{14}\ge\frac{14^2}{14}=14$ (2)
Din(1) si (2) rezulta inegalitatea de demonstrat.

a)Folosind inegalitatea lui Cauchy-Schwarz obtinem $(a^3+b^3)(a+b)\ge(a^2+b^2)^2\Rightarrow \frac{a^3+b^3}{a^2+b^2}\ge\frac{a^2+b^2}{a+b}$ (1)
Din ipoteza avem $\frac{a^2+b^2}{a+b}\ge 1$ (2)
Din (1) si (2) obtinem ca $\frac{a^3+b^3}{a^2+b^2}\ge 1\Rightarrow a^3+b^3\ge a^2+b^2$

Fie $x_0$ radacina comuna a polinoamelor $P(x)$ si $P(P(P(x)))$.Atunci avem:
$P(x_0)=0$ si $P(P(P(x_0)))=0$,deci $P(P(0))=0$. Cum $P(0)=b$ inseamna ca $P(b)=0\Rightarrow b^2+ab+b=0\Rightarrow b(b+a+1)=0\Rightarrow P(0)P(1)=0$, ceea ce trebuia demonstrat.