Page 43 - RMGO 5

Some identities and inequalities in triangle 43

2
8R − 7r X 1 2R − r
4. ≤ ≤ ;
2 3
2
16s r (b + c − a) (a + c − b) 2 4sr 2
X a
5. ≥ 3,
b + c − a
that is Problem M 7 from Kvant;
6. abc ≥ (a + b − c)(b + c − a)(c + a − b);

X a(b + c − a)
7. ≥ 3;
(a − b + c)(a + b − c)
8R − 7r X b + c − a (2R − r) 2
8. ≤ ≤ ;
2sr (a − b + c)(a + b − c) 2sr 2
2
2
2
2
4r(4R + 6Rr − r ) X (b + c − a)(a − b + c) 4r(5R + 3Rr + r )
9. ≤ ≤ ;
sR c sR
9 X a 9R
10. ≤ ≤
2s (a − b + c)(a + b − c) 4sr
(another refinement of Euler’s inequality);
3r X a 2 3
11. ≤ ≤
R (a − b + c)(a + b − c) 2
(another refinement of Euler’s inequality);

X a 2 4s(R − r)
12. ≥ ≥ 2s;
b + c − a R
2
4R − 5r X (b + c − a) 2 4R − 8Rr + 3r 2
13. ≤ ≤ ;
r (a − b + c)(a + b − c) r 2
2
2R − r X ab R − Rr + r 2
14. ≤ ≤ ;
r (b + c − a)(a − b + c) r 2
2
2
2(2R − r) X b + c − a 2(R − Rr + r )
15. ≤ ≤ ;
R a Rr
6r X (b + c − a)(a − b + c) 2R − r
16. ≤ ≤ .
R ab r
Proof. In Theorem 1 we use the Euler’s Inequality R ≥ 2r and the Gerretsen’s
Inequalities
2
2
2
2
16Rr − 5r ≤ s ≤ 4R + 4Rr + 3r .
We proof only the inequality 5), the another inequalities having similar proofs:
X a 2R − r 4r − r
= ≥ = 3.
b + c − a r r
This is a new proof of Problem M 7, Kvant.