Prove $b^2-c^2\leq2a^2$ in $\triangle ABC$

[Albert1]

$\triangle ABC$

$(1): \overline{AB}=c,\overline{BC}=a\,\, and \,\,\overline{CA}=b$

$(2)\angle B=30^o$

prove:

$b^2-c^2\leq 2a^2$

 

[Albert1]

$\triangle ABC$

$(1): \overline{AB}=c,\overline{BC}=a\,\, and \,\,\overline{CA}=b$

$(2)\angle B=30^o$

prove:

$b^2-c^2\leq 2a^2$

using law of cosine

Spoiler

$b^2=a^2+c^2-2ac\,cos\,B$
$\therefore b^2-c^2-2a^2=-a^2-\sqrt 3 \,ac\leq 0$
and the proof is done