Với c≥0,d≤0" role="presentation" style="font-size: 13.696px; display: inline; word-spacing: 0px; position: relative;">c≥0,d≤0 hoặc c≤0,d≥0" role="presentation" style="font-size: 13.696px; display: inline; word-spacing: 0px; position: relative;">c≤0,d≥0. Ta có cd≤0(1)" role="presentation" style="font-size: 13.696px; display: inline; word-spacing: 0px; position: relative;">cd≤0(1)Với c≤0,d≤0" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">c≤0,d≤0. Ta có c+d≤0" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">c+d≤0 ( ko thể xảy ra )
Với c≥0,d≥0" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">c≥0,d≥0. Áp dụng bđt AM−GM" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">AM−GM : cd≤(c+d)24=94(2)" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">cd≤(c+d)24=94(2)
Từ (1),(2)⇒cd≤94" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">(1),(2)⇒cd≤94 ∀cd" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">∀cd ∈R" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">∈R
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Áp dụng bđt bunhia, ta có (ac+bd)2≤(a2+b2)(c2+d2)⇔|ac+bd|≤c2+d2" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">(ac+bd)2≤(a2+b2)(c2+d2)⇔|ac+bd|≤c2+d2−−−−−−√
Ta lại có ac+bd≤|ac+bd|" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">ac+bd≤|ac+bd|
⇒ac+bd≤c2+d2" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">⇒ac+bd≤c2+d2−−−−−−√
⇔VT≤c2+d2+cd=9−2cd+cd=322.9−2cd322+cd" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">⇔VT≤c2+d2−−−−−−√+cd=9−2cd−−−−−−√+cd=32√2−−−−√.9−2cd−−−−−−√32√2−−−√+cd
≤322+9−2cd3222+cd=324+9−2cd32+cd" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">≤32√2+9−2cd32√22+cd=32√4+9−2cd32√+cd
=924+32−232.cd≤924+32−232.94" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">=92√4+32√−232√.cd≤92√4+32√−232√.94
VT≤9+624" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">VT≤9+62√4 (đpcm)
Dấu "="" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">"=" xảy ra ⇔a=b=22,c=d=32" role="presentation" style="font-size: 13.696px; position: relative;">⇔a=b=2√2,c=d=32