(1/2)三题:1)a(ac+bd)2 2)已知ad≠bc证明(a2+b2)(c2+d2)>(ac+bd)23)比较b/a和(b+m)/(a+m) (a,b,m∈R+)

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(1/2)三题:1)a(ac+bd)2 2)已知ad≠bc证明(a2+b2)(c2+d2)>(ac+bd)23)比较b/a和(b+m)/(a+m) (a,b,m∈R+)
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(1/2)三题:1)a(ac+bd)2 2)已知ad≠bc证明(a2+b2)(c2+d2)>(ac+bd)23)比较b/a和(b+m)/(a+m) (a,b,m∈R+)
(1/2)三题:1)a(ac+bd)2
2)已知ad≠bc证明(a2+b2)(c2+d2)>(ac+bd)2
3)比较b/a和(b+m)/(a+m) (a,b,m∈R+)

(1/2)三题:1)a(ac+bd)2 2)已知ad≠bc证明(a2+b2)(c2+d2)>(ac+bd)23)比较b/a和(b+m)/(a+m) (a,b,m∈R+)
(1)构造方程ax^2+bx+c=0,x=1时有a+b+c=0,所以x=1是该方程的解.△=b^2-4ac≥0
(2)(a^2+b^2)(c^2+d^2)-(ac+bd)^2=a^2d^2+b^2c^2-2abcd=(ab-bc)^2≥0,原不等式成立.
(3)当a>b时,b/a<(b+m)/(a+m)

(a^2+b^2)(c^2+d^2) (a,b,c,d∈R)
=a^2·c^2 +b^2·d^2+a^2·d^2+b^2·c^2
=a^2·c^2 +2abcd+b^2·d^2+a^2·d^2-2abcd+b^2·c^2
=(ac+bd)^2+(ad-bc)^2
≥(ac+bd)^2,
写不过啊!