min/max of ((a+b+c)^2)/bc

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Source: Bulgaria 1978 P4

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#1 h Jun 17, 2021, 5:55 PM • 3 Y

Y by Mango247, Mango247, Mango247

Find the greatest possible real value of $S$ and smallest possible value of $T$ such that for every triangle with sides $a,b,c$ $(a\le b\le c)$ to be true the inequalities:
$S\le\frac{(a+b+c)^2}{bc}\le T.$

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#2 h Jun 18, 2021, 12:53 PM

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Write it as $\frac{(a+(b+c))^2}{bc}$
Now use $A.M$ $\ge$ $G.M$
So $b+c\ge2\sqrt{bc}$
Now our expression is $\frac{(a^2+(b+c)^2+2a(b+c)}{bc}$
Which is equivalent to 4+ $\frac{(a^2+2\sqrt{2}a)}{bc}$
Now since $b+c$ > $a$
So $\frac{(a^2+2\sqrt{2}a)}{bc}$ < $\frac{((b+c)^2+2\sqrt{2}(b+c))}{bc}$
Which is $4+8+4=16-1=15$
Hence, $T=15$

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