Prove that $x^2 + 3 > 2x + 1$ for all values of $x$

Question

how would I go about proving this:

$x^2 + 3 > 2x + 1$

I know that I have to complete the square but do I bring everything to one side and convert it to an equation or do I simply complete the first side? Sorry if this seems nooby

move everything to the left and then complete the square. – paw88789 Sep 21 '17 at 19:38 — paw88789, Sep 21 '17 at 19:38

score 4 · Accepted Answer · answered Sep 21 '17 at 19:41

4

Hint:

Adding to both sides $-2x-1$ it is equivalent to $$ x^2-2x+2=(x-1)^2+1>0 $$

answered Sep 21 '17 at 19:41

Emilio Novati

62,675

score 4 · Answer 2 · answered Sep 21 '17 at 19:42

4

$x^2+3>2x+1 \Leftrightarrow x^2-2x+1+1>0\Leftrightarrow (x-1)^2+1>0$

Which is true for every x.

answered Sep 21 '17 at 19:42

Cornman

11,065
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What is wrong with my answer so you have to downvote it? – Cornman Sep 21 '17 at 19:47

score 2 · Answer 3 · answered Sep 21 '17 at 19:43

HINT:

Take everything over to the left hand side and then complete there square.

Example: Show that $x^2 > 2x-6$ for all $x$:

$$x^2 > 2x - 6 \iff x^2 - 2x + 6 > 0 \iff (x-1)^2 +5 > 0$$ Since $(x-1)^2 \ge 0$ for all $x \in \mathbb R$, we have $(x-1)^2 + 5 \ge 5$. Hence $(x-1)^2 + 5 > 0$, as required.

score 2 · Answer 4 · answered Sep 21 '17 at 20:10

With Analysis:

Just to show there isn't a single way to prove inequalities.

The tangent to the parabola with equation $y=x^2+3$ at point $A=(1,4)$ has equation $y=2(x-1)+4=2x+2$.

Now the function is convex, i.e. its graph is above each of its tangents, so for all $x$, $$x^2+3\ge 2x+2~(>2x+1).$$

Prove that $x^2 + 3 > 2x + 1$ for all values of $x$

4 Answers4