More trigonometry practice


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.)


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

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