Strengthened inequalities

Question

I have recently come across the concept of strengthening an inequality. What I can't understand is why under the same constraints both the strengthened and the original version work. The thing I can't get is how can both of them achieve their equality cases because for instance if both $A \ge 2B$ and $A\ge (2+\sqrt 3)B$ hold I can't see how the first one achieves its equality case.

Answer 1

A "stronger" inequality implies the "weaker" one directly, while the converse is not true in general.

For example, in non-negative numbers $\,\dfrac{x+y}{2} \ge \sqrt{xy}\,$ is a stronger inequality than $\,\sqrt{\dfrac{x^2+y^2}{2}} \ge \sqrt{xy}\,$ because for all non-negative $\,x,y\,$ it holds that $\,\sqrt{\dfrac{x^2+y^2}{2}} \ge \dfrac{x+y}{2}\,$ by the RMS-AM inequality, and therefore $\,\dfrac{x+y}{2} \ge \sqrt{xy} \implies \sqrt{\dfrac{x^2+y^2}{2}} \ge \dfrac{x+y}{2} \ge \sqrt{xy}\,$.

In the equality case $\,x=y\,$ they all match of course $\,\sqrt{\dfrac{x^2+x^2}{2}} = \dfrac{x+x}{2} = \sqrt{x\,x} = x\,$, but in all other cases the inequalities are strict $\,\sqrt{\dfrac{x^2+y^2}{2}} \gt \dfrac{x+y}{2} \gt \sqrt{xy}\,$, so it makes sense to call the "closest matching" inequality "stronger".

Answer 2

Let consider

• $A\ge2B$

• $A\ge(2+\sqrt 3)B$

and not that for $B=0$ both reduce to $A\ge 0$ otherwise we have two cases

• $B>0$

$$A\ge(2+\sqrt 3)B \implies A\ge2B$$

• $B<0$

$$ A\ge2B \implies A\ge(2+\sqrt 3)B $$

Answer 3

Equality doesn't have to hold in an inequality!

As a basic example, we can say $2 \geq 1$. But we won't ever have $2 = 1$. We can even "strengthen" this inequality: $2 \geq 1.9 = (1 + .09) \cdot 1$. We still never have $2 = 1$ or $2 = 1.9$.

As a more complicated example of non-equalities, we have $x^2 + 1 \geq x$ for all real $x$. In fact, $x^2 + 1$ is strictly greater than $x$ for all real $x$, so we never get $x^2 + 1 = x$.