elementary inequality proof

I am working with a howework inquiry, attempting to confirm the following:

$$5a+b > 4\sqrt{ab},$$

where $a$ and also $b$ declare actual numbers.

I've attempted increasing expression by $\sqrt{ab}$, making even both sides of the formula until now. In both instances, after re - factoring, I can not end that inequality holds. Can a person aim me in the appropriate instructions?

2022-07-25 16:44:12
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Answers: 3

For any kind of actual $x$ and also $y$ you have $$2xy=x^2+y^2-(x-y)^2$$ taking $x=2\sqrt{a}$ and also $y=\sqrt{b}$ you get $$2(2\sqrt{a}\sqrt{b})=4 a+b-(2\sqrt{a}-\sqrt{b})^2\leq 4a+b < 5a+b$$ the only continuing to be instance (i.e. $a=0$) is unimportant.

2022-07-25 19:41:15

In general you have to be a bit careful about squaring inequalities, but here there’s no problem: since we’re assuming that $a$ and $b$ are positive, $5a+b$ is positive, and therefore $5a+b>4\sqrt{ab}$ if and only if $(5a+b)^2>16ab$, i.e., if and only if $25a^2-6ab+b^2>0$. Now think of $b^2-6ab+25a^2$ as a quadratic in $b$ and complete the square: $b^2-6ab+25a^2=(b-3a)^2+16a^2$. Clearly $(b-3a)^2\ge 0$, and since $a$ is positive, $a^2>0$, so $(b-3)^2+16a^2>0$.

Retracing our steps, we see that $(5a+b)^2>16ab$ and hence that $5a+b=\sqrt{(5a+b)^2}>4\sqrt{ab}$.

2022-07-25 19:40:54

Let $a=x^2$ and $b=y^2$, where $x$ and $y$ are positive. We want to show that $5x^2-4xy+y^2 \gt 0$. This is clear from the fact that already $4x^2-4xy+y^2=(2x-y)^2 \ge 0$.

Alternately, note that $5x^2-4xy+y^2=x^2((y/x)^2-4(y/x)+5)$. Let $t=y/x$, and use the fact that $t^2-4t+5$ is positive at (say) $t=0$, and is never $0$, so it is always positive.

2022-07-25 19:39:40