## Wednesday, August 19, 2015

### Prove $35√55+55√77+77√35+35√77+55√35+77√55\gt 2310$.

Prove $35\sqrt{55}+55\sqrt{77}+77\sqrt{35}+35\sqrt{77}+55\sqrt{35}+77\sqrt{55}\gt 2310$.

The first idea that might sprang to your mind whenever you see square root terms and the inequality is you want to squaring both sides of the inequality and you are expected to finally prove a figure is larger than another figure and you are hence done.

But, in this instance, even after we grouped the like terms, we still have a total of $3$ square roots on the left, and we have to square more than one time in order to clearing all of the square roots and get eventually only numbers on our site.

Still, that is a workable approach though.

We first assume the given inequality holds:

$35\sqrt{55}+55\sqrt{77}+77\sqrt{35}+35\sqrt{77}+55\sqrt{35}+77\sqrt{55}\gt 2310$

Grouping the like terms, we have:

$\sqrt{55}(35+77)+\sqrt{77}(55+35)+\sqrt{35}(77+55)\gt 2310$

$112\sqrt{55}+90\sqrt{77}+132\sqrt{35}\gt 2310$

Now, we can square both sides of the inequality (only after we moved any one of the term to the right side):

$112\sqrt{55}+90\sqrt{77}\gt 2310-132\sqrt{35}$

$(112\sqrt{55}+90\sqrt{77})^2\gt (2310-132\sqrt{35})^2$

$112^2(55)+2(112)(90)\sqrt{(55)(77)}+90^2(77)\gt 2310^2-2(2310)132\sqrt{35}+132^2(35)$

$1313620+2(112)(90)\sqrt{(55)(77)}\gt 5945940-2(2310)132\sqrt{35}$

$2(112)(90)\sqrt{(55)(77)}\gt 4632320-2(2310)132\sqrt{35}$

$2^4(1260)\sqrt{(55)(77)}\gt 2^4(289520)-2^4(38115)\sqrt{35}$

$1260\sqrt{(55)(77)}\gt 289520-38115\sqrt{35}$

Squaring both sides again yields:

$(1260\sqrt{(55)(77)})^2\gt (289520-38115\sqrt{35})^2$

$1260^2(55)(77)\gt 289520^2-2(289520)38115\sqrt{35}+38115^2(32)$

$5^2(317520)(11)(77)\gt 5^2(57904)^2-5^2(2(441402192))\sqrt{35}+5^2(7623^2(32))$

$(317520)(11)(77)\gt (57904)^2-(2(441402192))\sqrt{35}+(7623^2(32))$

$(2(441402192))\sqrt{35}\gt (57904)^2-(317520)(11)(77)+(7623^2(32))$

$\sqrt{35}\gt \dfrac{4943457904}{2(441402192)}$

$\sqrt{35}\gt \dfrac{52111}{9306}$

Squaring again gives:

$35>31.35687091$

which is certainly correct.

Phew, it is necessarily daunting with work with so many large numbers which the calculator couldn't handle it, and we have to rely on wolfram (yes, I used wolfram to compute the sum) and also to factor out the common factors along the way).

There has to be another easier way of attacking!

Of course there is! I will let you to have some fun thinking, and I will show you the heuristic method of solving in my next blog post.