Another method to prove $7 ≥ \sqrt 2+\sqrt 5 + \sqrt {11}$

Show with proof which of these two values is smaller:

$7$, or $\sqrt 2+\sqrt 5 + \sqrt {11}$

In my (few) previous blog post (Which is greater), I mentioned of how I proved for $\sqrt 2+\sqrt 5 + \sqrt {11}$ is smaller than $7$.

But that doesn't mean that solution is the only way out to prove for that kind of problem.

If you have not seen what I have observed, you could still be able to prove it, provided you know how to square any given math expression to get rid of the square root terms.

Without knowing which is greater in the beginning, you are safe to assume either way.

Let's say we assume that $\sqrt 2+\sqrt 5 + \sqrt {11}> 7$ is true.

We then keep squaring both sides of the inequality until we obtain numerical value on both sides of the inequality sign:

$\sqrt 2+\sqrt {11}> 7-\sqrt 5$

$(\sqrt 2+\sqrt {11})^2> (7-\sqrt 5)^2$

$2+11+2\sqrt {2(11)}> 49+5-2(7)\sqrt 5$

$2\sqrt {2(11)}+14\sqrt 5> 41$

$(2\sqrt {2(11)}+14\sqrt 5)^2> (41)^2$

$4(2(11))+14^2(5)+2(2)(14)\sqrt {2(11)(5)}> (41)^2$

$2(2)(14)\sqrt {2(11)(5)}> 613$

$(2(2)(14)\sqrt {2(11)(5)})^2> (613)^2$

$344,960> 375,769$

This is pure wrong therefore the converse of our previous assumption must be true, i.e. we can conclude by now that

$\sqrt 2+\sqrt 5 + \sqrt {11}< 7$

As you can probably tell, this method is not "grant" nor "superior" but you must know that in solving any given mathematical problem, solving it is what matters and you must not look down on a seemingly "weak" solution.

Instead, all schools around the world are continually seeking ways to improve their teaching methodology because we are eager and want to produce critical thinkers to push our country and economy moving forward. This is good news, as through the improved version of the teaching methods, we can always learn more, so we will never be satisfied with where we are today.