Let
Tn(x)=n∑k=0bkcos(kx)Tn(x)=n∑k=0bkcos(kx)
with b2k=b2k+1=14k(2kk)(k≥0).b2k=b2k+1=14k(2kk)(k≥0).
In 1958, Vietoris proved that Tn(x)>0(n≥1;x∈(0,π)).Tn(x)>0(n≥1;x∈(0,π)).
We offer the following improvement of this result: The inequalities Tn(x)≥c0+c1x+c2x2>0(ck∈R,k=0,1,2)Tn(x)≥c0+c1x+c2x2>0(ck∈R,k=0,1,2)
hold for all n≥1n≥1 and x∈(0,π)x∈(0,π) if and only if c0=π2c2,c1=−2πc2,0<c2≤α,c0=π2c2,c1=−2πc2,0<c2≤α,
where α=min0≤t<πT6(t)(t−π)2=0.12290….