20xx高考數(shù)學(xué)文人教a版一輪復(fù)習(xí)學(xué)案：選修4—5-第2課時(shí)-不等式的證明-【含解析】(參考版)
2025-04-03 03:29本頁面
【正文】 1a+c+a+b2abc=8,當(dāng)且僅當(dāng)a=b=c=13時(shí),等號成立.(2)ab+c+ba+c+ca+b=a+b+cb+c1+a+b+ca+c1+a+b+ca+b1=12[(b+c)+(a+c)+(a+b)]1b+c+1a+c+1a+b3≥12b+ca+bc≥2bca1cc=b+ca1]2=(a+b+c+1)2=9,∴a2+(b1)2+(c+2)2≥3,當(dāng)且僅當(dāng)a=b1=c+2,a+b+c=2,即a=1,b=2,c=1時(shí),等號成立.∴a2+b2+c22b+4c+2≥0.對點(diǎn)訓(xùn)練6證明(1)1a11b11c1=1aa1+(b1)當(dāng)x0時(shí),得2x+1≥4,此時(shí)x≤32.所以不等式的解集為xx≥52或x≤32.(2)證明f(x)+f1x=|x+1|+1x1,由絕對值三角不等式,得|x+1|+1x1≥x+1x,又x,1x同號,所以x+1x=|x|+1x,由基本不等式,得|x|+1x≥2,當(dāng)且僅當(dāng)|x|=1時(shí),等號成立.所以f(x)+f1x≥2.對點(diǎn)訓(xùn)練4 (1)解當(dāng)m=1時(shí),f(x)=|x1|+x+12=2x12,x1,32,12≤x≤1,2x+12,x12.于是原不等式可化為x1,2x12≤3或12≤x≤1,32≤3或x12,2x+12≤3,解得1x≤74或12≤x≤1或54≤x12.所以不等式f(x)≤3的解集為54,74.(2)證明由已知得f(x)+1m(m+1)=|xm|+x+1m+1+1m(m+1)≥xmx+1m+1+1m(m+1)=m+1m+1+1m(m+1)=m+1m+1+1m1m+1=m+1m≥2,當(dāng)且僅當(dāng)m=1時(shí),.例5(2)證明|x+2||x2|≤|(x+2)(x2)|=4,由于0y1,則1y+11y=1y+11y[y+(1y)]=2+1yy+y1y≥2+2=4,當(dāng)且僅當(dāng)1yy=y1y,即y=12時(shí)取等號.綜上,|x+2||x2|≤1y+11y.對點(diǎn)訓(xùn)練5(1)解當(dāng)a=b=c=2時(shí),f(x)=|x2|+|x+2|+2,所以f(x)8?x≤2,22x8,或2x2,68,或x≥2,2x+28,所以不等式的解集為{x|3x3}.(2)證明因?yàn)閍0,b0,c0,所以f(x)=|ax|+|x+b|+c≥|ax+x+b|+c=|a+b|+c=a+b+c.因?yàn)閒(x)的最小值為1,所以a+b+c=1,所以(a+b+c)2=a2+b2
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