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I would appreciate if someone could help me with this problem.

Assume $a$, $b$ and $c$ are real positive values where $a^2 + b^2 +c^2 =1$.

Is it possible to prove that $ab+ac+bc\leq 1$?

obviously special cases e.g. $a=b=c=1/3$ hole for the case of equality but I am interested in a general solution.

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    That is just an instance of the rearrangement inequality. – 2017-02-02
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    Hint: prove that $ab+bc+ca \le a^2+b^2+c^2$ for $\forall a,b,c \ge 0\,$. – 2017-02-02
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    Further hint: $$0\leq (a-b)^2+(a-c)^2+(b-c)^2=\ldots$$ – 2017-02-02
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    many thanks for the hints! – 2017-02-02

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$$1-ab-ac-bc=a^2+b^2+c^2-ab-ac-bc=\sum_{cyc}(a^2-ab)=$$ $$=\frac{1}{2}\sum_{cyc}(2a^2-2ab)=\frac{1}{2}(a^2+b^2-2ab)=\frac{1}{2}\sum_{cyc}(a-b)^2\geq0$$