More trigonometry practice

Problem.

Let ABC be a triangle and $h_{a}$ be the altitude through A. Prove that

$(b+c)^{2} \geq a^{2}+4(h_{a})^{2}$ .

(As usual, a, b, c denote the sides BC, CA, AB respectively.)

Proof.

The given inequality is equivalent to $(b+c)^{2}-a^{2} \geq 4(h_{a})^{2}=\frac{16\triangle^{2}}{a^{2}}$ .

where $\triangle$ is the area of triangle ABC. Using the identity

$16\triangle^{2}=[(b+c)^{2}-a^{2}][a^{2}-(b-c)^{2}]$ , we see that the inequality to be proved is

$a^{2}-(b-c)^{2} \leq a^{2}$ (here we use $a<b+c$ ), which is true. Observe that equality holds iff $b=c$ . QED.

More later,

Nalin Pithwa

This entry was written by nkpithwa, posted on July 23, 2015 at 11:06 am, filed under Cnennai Math Institute Entrance Exam, IITJEE Mains, ISI Kolkatta Entrance Exam, RMO, Trigonometry and tagged area of a triangle formula. Bookmark the permalink. Follow any comments here with the RSS feed for this post. Post a comment or leave a trackback: Trackback URL.

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