Question

In ∆PQR, if S is any point on the side QR, show that PQ + QR + RP > 2PS

Answer 1

in ∆PQS
PQ+QS≥PS (1)(sum of two sides of tri is greater than third side)

now, in ∆PSR
PR +RS ≥ PS (2) (sum of two sides of a triangle is greater than third side)
add equation (1) and (2)
PR+SR≥PS
+ PQ+QS≥PS
=PR+SR+PQ+QS ≥PS + PS
=PR+PQ+ QR ≥ 2PS
Hence proved
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Answer 2

now, in ∆PSR

PR +RS ≥ PS (2) (sum of two sides of a triangle is greater than third side)

add equation (1) and (2)

PR+SR≥PS

+ PQ+QS≥PS

=PR+SR+PQ+QS ≥PS + PS

=PR+PQ+ QR ≥ 2PS