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Prove that,In any tetrahedron,ther is a verticle such that all of the angles connected with that verticle are less than right angle.

yep, my counter example was wrong. Here's my solution :

The 4 points are not coplaner. There are 4 triangles, each can have atmost one non-acute angle. So there can be atmost 4 non-acute angles in total. So if we assume there is no such vertex as stated above, then each triangle and each vertex should have exactly one non-acute angle.
If a vertex has three angles $x,y,z$, then any 2 of them must be greater then the third (prove this yourself). Let $ABCD$ be the tetrahedron and let $AB$ be a edge with the largest length. Then $\angle ACB$ and $\angle ADB $ must be non-acute. $\angle CAB ,\angle BAD ,\angle ABD,\angle CBA$ are acute. So $ \angle DAC$ and $\angle DBC$ are both non-acute. But $\angle DCA+ \angle DCB >\angle ACB$ and $\angle CDA +\angle CDB> \angle ADB$. So,

$$\angle DCA+ \angle DCB+\angle CDA +\angle CDB>\angle ACB+\angle ADB\geq 90^\circ+90^\circ=180^\circ$$
Again,
$$\angle DCA+ \angle DCB+\angle CDA +\angle CDB= (180^\circ-\angle DAC)+(180^\circ-\angle DBC)\leq 90^\circ+90^\circ=180^\circ$$

This gives a contradiction. So there must be a vertex as stated above.

I don't know how to solve it but this problem is taken from 'Bigganchinta' and Nafis Tiham Vhai has given this problem.

Thanks for the solution.I was thinking wrong about solving this problem.