Using your Head is Permitted

April 2015 riddle

This month, as last month, the riddle is a follow-up to Qiyang Lu's February challenge. Also as last month, both the riddle and its solution are due to Radu-Alexandru Todor.

Unlike last month, this month please do send in your solutions. Also unlike last month, this time we are dealing with non-linear inequalities.

Specifically, prove that for any quadrilateral with side lengths a≥b≥c≥d and diagonal lengths e and f the following holds:

(e+f)2 ≤ min((a+b)2+4cd,(c+d)2+4ab).

This month's bonus question (for an asterisk next to your name):

Prove that for every t>0 and any quadrilateral with sides and diagonals as described above the following holds:

e+f ≤ a+b+(t+1/t-2)c+td.

Incidentally, I am not presently aware of any general characterisation for the tuples (α,β,γ,δ) for which

e+f ≤ αa+βb+γc+δd

holds for every quadrilateral, but would be most interested to see such a characterisation with a proof for it if any reader is able to find one.

List of solvers:

Li Li (*) (5 April 16:18)
Zhengpeng Wu (*) (7 April 00:57)

Elegant and original solutions can be submitted to the puzzlemaster at riddlesbrand.scso.com. Names of solvers will be posted on this page. Notify if you don't want your name to be mentioned.

The solution will be published at the end of the month.

Enjoy!

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