a.Xét t≥a và g(m)=t∫a[mx−f(x)]2dx,m∈R
⇒g(m)=(t∫ax2dx)m2−2(t∫axf(x)dx)m+t∫a[f(x)]2dx≥0,∀m∈R
⇒Δ′=(t∫axf(x)dx)2−(t∫ax2dx)(t∫a[f(x)]2dx)≤0
⇒(t∫axf(x)dx)2≤(t∫ax2dx)(t∫a[f(x)]2dx)≤(t∫ax2dx)(t∫ax2dx)
(Do giả thiết: t∫a[f(x)]2dx≤t∫ax2dx)
⇒t∫axf(x)dx≤t∫ax2dx
⇒F(t)=t∫ax[x−f(x)]dx≥0
⇒F′(t)=t[t−f(t)]
Xét: b∫a[x−f(x)]dx=b∫a1x[x(x−f(x))]dx
=b∫a1xF′(x)dx,∀b≥a
Đặt: {u=1xdv=F′(x)dx
⇒{du=−dxx2v=F(x)
⇒b∫a[x−f(x)]dx=[1xF(x)]ba+b∫aF(x)x2dx
=1bF(b)−1aF(a)+b∫aF(x)x2dx
=1bF(b)+b∫aF(x)x2dx≥0
(vì F(a)=0)
(vì F(b),F(x)≥0)
⇒b∫a[x−f(x)]dx≥0
⇒b∫af(x)dx≤b∫axdx
b.Ta có: b∫a[t√f(x)−1√f(x)]2dx≥0,∀t∈R
⇒b∫a[t2f(x)−2t+1f(x)]dx≥0,∀t∈R
⇒(b∫af(x)dx)t2−2(b−a)t+b∫adxf(x)≥0,∀t∈R
⇒Δ′=(b−a)2−(b∫af(x)dx)(b∫adxf(x))≤0
⇒(b∫af(x)dx)(b∫adxf(x))≥(b−a)2
⇒ (ĐPCM)