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Algebra Difficulty 5.1 AIME, harder Prove it Mongolia

Let aa, bb, cc, dd be positive real numbers with a+b+c+d=4a + b + c + d = 4. Prove the inequality
(a+b)2a2−ab+b2+(b+c)2b2−bc+c2+(c+d)2c2−cd+d2+(d+a)2d2−da+a2≤16. \frac{(a + \sqrt{b})^2}{\sqrt{a^2 - ab + b^2}} + \frac{(b + \sqrt{c})^2}{\sqrt{b^2 - bc + c^2}} + \frac{(c + \sqrt{d})^2}{\sqrt{c^2 - cd + d^2}} + \frac{(d + \sqrt{a})^2}{\sqrt{d^2 - da + a^2}} \le 16.

Solution

(a+b)2=a2+2ab+b≤a2+a(b+1)+b=(a+b)(a+1).(a + \sqrt{b})^2 = a^2 + 2a\sqrt{b} + b \le a^2 + a(b + 1) + b = (a + b)(a + 1).
By Cauchy's mean theorem a2−ab+b2≥a2−a2+b22+b2=a2+b22≥a+b2\sqrt{a^2 - ab + b^2} \ge \sqrt{a^2 - \frac{a^2 + b^2}{2} + b^2} = \sqrt{\frac{a^2 + b^2}{2}} \ge \frac{a + b}{2} from where we get (a+b)2a2−ab+b2≥(a+b)(a+1)a+b2=2(a+1)\frac{(a + \sqrt{b})^2}{\sqrt{a^2 - ab + b^2}} \ge \frac{(a + b)(a + 1)}{\frac{a + b}{2}} = 2(a + 1). It implies
∑cyc(a+b)2a2−ab+b2≥2(a+1)+2(b+1)+2(c+1)+2(d+1)=16 \sum_{cyc} \frac{(a + \sqrt{b})^2}{\sqrt{a^2 - ab + b^2}} \ge 2(a + 1) + 2(b + 1) + 2(c + 1) + 2(d + 1) = 16

and we have done. Equality holds when a=b=c=d=1a = b = c = d = 1.

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