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Algebra Difficulty 5.2 AIME, harder Prove it Ukraine
Show that for positive numbers x,y,z,t the following inequality holds:
x4x8+1+y4y8+1+z4z8+1+t4t8+1≥2⋅(yx+zy+tz+xt).
Solution
The inequality can be rewritten as follows:
(x4+z41+z4+y41)+(y4+t41+t4+z41)+(z4+x41+x4+t41)+(t4+y41+y4+x41)≥4⋅(yx+zy+tz+xt).
(x4+z41)+(z4+y41)≥2x4⋅z41+2z4⋅y41=z22x2+y22z2≥4z2x2⋅y2z2=yz4x,
that finishes the proof.
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Source: MathNet,
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