Problem:
Jika $a,b,c$ bilangan real positif, buktikan bahwa :
$$frac{a}{b^2 + c^2 + d^2} +frac{ b}{a^2 + c^2 + d^2} + frac{c}{a^2 + b^2 + d^2} + frac{d}{a^2 + b^2 + c^2} ge frac{4}{a+b+c+d}$$
Jika $a,b,c$ bilangan real positif, buktikan bahwa :
$$frac{a}{b^2 + c^2 + d^2} +frac{ b}{a^2 + c^2 + d^2} + frac{c}{a^2 + b^2 + d^2} + frac{d}{a^2 + b^2 + c^2} ge frac{4}{a+b+c+d}$$
Solusi:
Dengan mengaplikasikan CS Inequality, didapat:
$$(a+b+c+d)left(sum_{cyc}frac{a}{b^2+c^2+d^2}right)geleft(sum_{cyc}sqrt{frac{a^2}{b^2+c^2+d^2}}right)^2$$
akan tetapi berdasarkan AM-GM, didapat:
$$sqrt{frac{b^2+c^2+d^2}{a^2}} le frac{1}{2}left(frac{b^2+c^2+d^2}{a^2}+1right)=frac{1}{2}left(frac{a^2+b^2+c^2+d^2}{a^2}right)$$
sehingga:
$$
left(sum_{cyc}sqrt{frac{a^2}{b^2+c^2+d^2}}right)^2geleft(sum_{cyc}frac{2a^2}{a^2+b^2+c^2+d^2}right)^2=2^2$$
dengan demikian
$$begin{aligned}(a+b+c+d)left(sum_{cyc}frac{a}{b^2+c^2+d^2}right)&ge4\
left(sum_{cyc}frac{a}{b^2+c^2+d^2}right)&gefrac{4}{(a+b+c+d)}end{aligned}$$