Cho 3 số thực dương phân biệt a,b,c:(0<a<b<c)
Gọi P=4(a+b+c),Q=2(ab+bc+ca),d=a2+b2+c2
Chứng minh: a<13(P4dQ2)<13(P4+dQ2)<c
Đặt {α=13(P4dQ2)β=13(P4+dQ2){α+β=P6α.β=19(P216d+Q2)
Ta có:
* dQ2=(a2+b2+c2)(ab+bc+ca)=12[(ab)2+(bc)2+(ca)2]>0 ( do a<b<c)
Bài toán có nghĩa và α<β(1)
α,β là hai nghiệm phân biệt của phương trình
f(t):t2P6t+19(P216d+Q2)=0(2)
* S2=α+β2=P12=a+b+c3a<S2<c  ( do a<b<c)    (3)
* f(a)=a2P6a+19(P216d+Q2)=a22a(a+b+c)3+ab+bc+ca3
      =a2ab+bcca3=frac(ab)(ac)3>0    ( do a<b<C)    (4)
* Tương tự f(c)=(ca)(cb)3>0   ( do a<b<c)      (5)
* Từ (2),(3),(4),(5) suy ra phương trình (1) có các nghiệm α,β thỏa mãn a<α<β<c. Đó là đpcm

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