My teacher told me that use Cauchy-Schwarz, but I still can’t prove the inequality.
I just tried taking it all apart and make $(2a+2b)^2$ because it has $8ab$ and I don't what to do after.
Can anyone give me some tips?
2026-04-07 09:55:22.1775555722
Prove that for all real numbers $a, b$ we will have inequality: $ab(a+2)(b+4)$+$5a^{2}$+$7b^2$+$10a$+$28b$+$29$ ≥$ 0$
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We want to show: $$5(a+1)^2+7(b+2)^2+a(a+2)\cdot b(b+4)-4\geqslant 0$$ This suggests using $x = a+1, y=b+2$, so we may write this equivalently as $$5x^2+7y^2+(x^2-1)(y^2-4)-4= x^2+6y^2+x^2y^2\geqslant 0$$ which is obvious.
P.S.: I am not sure if finding a suitable CS inequality is worth the effort here...