Nice problem due to sir dan sitaru!



  • If a,b,ca,b,c and dd are positive real numbers for which 3(adbc)=bd+ac0\sqrt3(ad-bc)=bd+ac\neq0.
    Prove that
    d(a+b3)c(ba3)>4βd(a+b\sqrt3)-c(b-a\sqrt3)\gt 4 \beta
    where β=abcd4\beta=\sqrt[4]{abcd}.



  • My solution:
    First,note that
    3(adbc)=ac+bdd(a3b)=c(a+b3)\sqrt3(ad-bc)=ac+bd\Rightarrow d(a\sqrt3-b)=c(a+b\sqrt3)
    that is

    (1): adbcad\succ bc .(2):a3ba\sqrt3\succ b .(3):abcdβ\sqrt {abcd}\succ \beta
    So,put this in mind,and write the desired inequality as folllows
    (d2+c2d)(a+b3)4β.\left(\frac{d^2+c^2}{d}\right)(a+b\sqrt3)\succ 4\beta.
    Now,using (1) and (2) to get
    abcd(d2+c2d)(a+b3)bd(d2+c2)(3+cd)(d2+c2d)(a+b3)bd(d2+c2)(ba+cd)a\succ\frac{bc}{d}\Rightarrow\left(\frac{d^2+c^2}{d}\right)(a+b\sqrt3)\succ\frac{b}{d}(d^2+c^2)\left(\sqrt3+\frac{c}{d}\right)\Rightarrow\left(\frac{d^2+c^2}{d}\right)(a+b\sqrt3)\succ \frac{b}{d}(d^2+c^2)\left(\frac{b}{a}+\frac{c}{d}\right) ..(4)
    Now,due to AM-GM inequality we shall obtain that the RhsRhs of (4) is greater than 4abcd4\sqrt{abcd},which ends the proof because of (3).And we are done!


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