Let a,b and c are sides of ΔABC Δ is area of ΔABC. s= Semi-perimetre of ΔABC 2s=a+b+c and Δ2=s(s−a)(s−b)(s−c) By apply property of AM and GM, we get
s−a+s−b+s−c
3
≥((s−a)(s−b)(s−c))1∕3
3s−2s
3
≥(
Δ2
s
)
1
3
⇒
s
3
≥(
Δ2
s
)
1
3
⇒
Δ2
s
≤
s3
27
⇒Δ2≤
s4
27
⇒s2≥3√3Δ ⇒(
a+b+c
2
)2≥3√3Δ ⇒(a+b+c2≥4×3√3Δ ⇒(a+b+c)2≥12√3Δ We know that, a2+b2+c2≥