Ta có: 0≤(a+b+c)2=(a2+b2+c2)+2(ab+bc+ca)" role="presentation" style="font-size: 13.696px; display: inline; word-spacing: 0px; position: relative;">0≤(a+b+c)2=(a2+b2+c2)+2(ab+bc+ca)0≤(a+b+c)2=(a2+b2+c2)+2(ab+bc+ca)⇒ab+bc+ca≥−12" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">⇒ab+bc+ca≥−12⇒ab+bc+ca≥−12Đẳng thức khi {a+b+c=0a2+b2+c2=1" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">{a+b+c=0a2+b2+c2=1{a+b+c=0a2+b2+c2=1Mặt khác : a2+b22" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">a2+b22a2+b22≥" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">≥≥ −ab⇒" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">−ab⇒−ab⇒ ab≥" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">ab≥ab≥ -a2+b22" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">a2+b22a2+b22⇒F=ab+bc+ca+bc≥−4−a2+c22" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">⇒F=ab+bc+ca+bc≥−4−a2+c22⇒F=ab+bc+ca+bc≥−4−a2+c22Ta có : a2+b2≤a2+b2+c2=1" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">a2+b2≤a2+b2+c2=1a2+b2≤a2+b2+c2=1⇒12(a2+c2)≥−12⇒F≥−1" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">⇒12(a2+c2)≥−12⇒F≥−1⇒12(a2+c2)≥−12⇒F≥−1Vậy min F=−1" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">F=−1F=−1 khi {c=0a+b=0" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">{c=0a+b=0{c=0a+b=0⇒..." role="presentation" style="font-size: 12.8px; display: inline; position: relative;">⇒...

Max:(3+22)x2+y2≥2(1+2)xy" role="presentation" style="font-size: 13.696px; display: inline; word-spacing: 0px; position: relative;">(3+22√)x2+y2≥2(1+2√)xy(3+22)x2+y2≥2(1+2)xy(3+22)x2+z2≥2(1+2)xz" role="presentation" style="font-size: 13.696px; display: inline; word-spacing: 0px; position: relative;">(3+22√)x2+z2≥2(1+2√)xz(3+22)x2+z2≥2(1+2)xz2(1+2)(y2+z2)≥4(1+2)yz" role="presentation" style="font-size: 13.696px; display: inline; word-spacing: 0px; position: relative;">2(1+2√)(y2+z2)≥4(1+2√)yz2(1+2)(y2+z2)≥4(1+2)yz⇒(3+22)(x2+y2+z2)≥2(1+2)(xy+2yz+zx)" role="presentation" style="font-size: 13.696px; display: inline; word-spacing: 0px; position: relative;">⇒(3+22√)(x2+y2+z2)≥2(1+2√)(xy+2yz+zx)⇒(3+22)(x2+y2+z2)≥2(1+2)(xy+2yz+zx)⇔3+22≥2(1+2)F⇒F≤1+22" role="presentation" style="font-size: 13.696px; word-spacing: 0px; position: relative;">⇔3+22√≥2(1+2√)F⇒F≤1+2√2Min:Ta có: 0≤(a+b+c)2=(a2+b2+c2)+2(ab+bc+ca)" role="presentation" style="font-size: 13.696px; display: inline; word-spacing: 0px; position: relative;">0≤(a+b+c)2=(a2+b2+c2)+2(ab+bc+ca)0≤(a+b+c)2=(a2+b2+c2)+2(ab+bc+ca)⇒ab+bc+ca≥−12" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">⇒ab+bc+ca≥−12⇒ab+bc+ca≥−12Đẳng thức khi {a+b+c=0a2+b2+c2=1" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">{a+b+c=0a2+b2+c2=1{a+b+c=0a2+b2+c2=1Mặt khác : a2+b22" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">a2+b22a2+b22≥" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">≥≥ −ab⇒" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">−ab⇒−ab⇒ ab≥" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">ab≥ab≥ -a2+b22" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">a2+b22a2+b22⇒F=ab+bc+ca+bc≥−4−a2+c22" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">⇒F=ab+bc+ca+bc≥−4−a2+c22⇒F=ab+bc+ca+bc≥−4−a2+c22Ta có : a2+b2≤a2+b2+c2=1" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">a2+b2≤a2+b2+c2=1a2+b2≤a2+b2+c2=1⇒12(a2+c2)≥−12⇒F≥−1" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">⇒12(a2+c2)≥−12⇒F≥−1⇒12(a2+c2)≥−12⇒F≥−1Vậy min F=−1" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">F=−1F=−1 khi {c=0a+b=0" role="presentation" style="font-size: 12.8px; display: inline; position: relative;">{c=0a+b=0{c=0a+b=0⇒..." role="presentation" style="font-size: 12.8px; display: inline; position: relative;">⇒...

http://toan.hoctainha.vn/Hoi-Dap/Cau-Hoi/131652/toan-10

Ko mất tính tổng quát, giả sử x2≥x1" role="presentation" style="font-size: 13.696px; display: inline; word-spacing: 0px; position: relative;">x2≥x1x2≥x1Ta có : x4x3=x3x2=x2x1≥1" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">x4x3=x3x2=x2x1≥1x4x3=x3x2=x2x1≥1⇒x4≥x3≥x2≥x1" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">⇒x4≥x3≥x2≥x1⇒x4≥x3≥x2≥x1Phương trình x2−3x+m=0" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">x2−3x+m=0x2−3x+m=0 có 2" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">22 nghiệm x2≥x1làx2=3+Δ12;x2=3−Δ12" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">x2≥x1làx2=3+Δ1−−−√2;x2=3−Δ1−−−√2x2≥x1làx2=3+Δ12;x2=3−Δ12 với Δ1=9−4m≥0" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">Δ1=9−4m≥0Δ1=9−4m≥0 hay m≤94" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">m≤94m≤94Phương trình x2−12x+n" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">x2−12x+nx2−12x+n có 2" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">22 nghiệm x4≥x3làx4=6+Δ2;x3=6−Δ2" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">x4≥x3làx4=6+Δ2−−−√;x3=6−Δ2−−−√x4≥x3làx4=6+Δ2;x3=6−Δ2 với đk Δ2=36−n≥0" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">Δ2=36−n≥0Δ2=36−n≥0 hay n≤36" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">n≤36n≤36∗x2x1=x4x3⇒3+Δ13−Δ1=6+Δ26−Δ2" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">∗x2x1=x4x3⇒3+Δ1−−−√3−Δ1−−−√=6+Δ2−−−√6−Δ2−−−√∗x2x1=x4x3⇒3+Δ13−Δ1=6+Δ26−Δ2Nhân chéo rút gọn đc 2Δ1=Δ2(1)" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">2Δ1−−−√=Δ2−−−√(1)2Δ1=Δ2(1) ∗x2x1=x3x2⇒x22=x1x3⇒(3+Δ12)2=(3−Δ12)(6−Δ2)(2)" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">∗x2x1=x3x2⇒x22=x1x3⇒(3+Δ1−−−√2)2=(3−Δ1−−−√2)(6−Δ2−−−√)(2)∗x2x1=x3x2⇒x22=x1x3⇒(3+Δ12)2=(3−Δ12)(6−Δ2)(2)Thế (1)" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">(1)(1) vào (2)" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">(2)(2), ta có (3+Δ12)2=(3−Δ12)(6−2Δ1)" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">(3+Δ1−−−√2)2=(3−Δ1−−−√2)(6−2Δ1−−−√)(3+Δ12)2=(3−Δ12)(6−2Δ1)⇒(3+Δ12)2=(3−Δ1)2" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">⇒(3+Δ1−−−√2)2=(3−Δ1−−−√)2⇒(3+Δ12)2=(3−Δ1)2⇒3+Δ1=2(3−Δ1)⇒Δ1=1⇒Δ1=1" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">⇒3+Δ1−−−√=2(3−Δ1−−−√)⇒Δ1−−−√=1⇒Δ1=1⇒3+Δ1=2(3−Δ1)⇒Δ1=1⇒Δ1=1⇒Δ2=2⇒Δ2=4" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">⇒Δ2−−−√=2⇒Δ2=4⇒Δ2=2⇒Δ2=4Ta có {Δ1=1Δ2=4⇒{9−4m=136−n=4⇔{m=2n=32" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">{Δ1=1Δ2=4⇒{9−4m=136−n=4⇔{m=2n=32{Δ1=1Δ2=4⇒{9−4m=136−n=4⇔{m=2n=32 (thõa đk)