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edit approved Dec 9, 2023 at 10:16
The Amplitwist
edit approved Dec 9, 2023 at 10:16
The Amplitwist
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I do not know how far these methods can go, but in both your examples the maximum can be computed easily and without calculus using some algebraic inequality techniques, the kind of "standard tricks" that is taught to contestants in high-school math Olympiads.

For the first inequality, AM-GM suffices: $$ xy^2 = 4 (x \cdot y/2 \cdot y/2) \leq 4 \left(\frac{x+y/2+y/2}{3}\right)^3 = \frac{4}{27} (x+y)^3 $$ (look also for weighted AM-GMweighted AM-GM).

For the second, use one of Maclaurin's inequalitiesMaclaurin's inequalities in three variables to get $$ (yz+yw+zw) \leq \frac{1}{3}\left(y+z+w\right)^2, $$ and then the inequality can be reduced to the previous one by setting $Y=y+z+w$.

This kind of tricks can work in simple cases, or where there is much symmetry in the variables (check Muirhead's inequalityMuirhead's inequality for instance for another highly-symmetric case); if this is not your case you may be out of luck though.

I do not know how far these methods can go, but in both your examples the maximum can be computed easily and without calculus using some algebraic inequality techniques, the kind of "standard tricks" that is taught to contestants in high-school math Olympiads.

For the first inequality, AM-GM suffices: $$ xy^2 = 4 (x \cdot y/2 \cdot y/2) \leq 4 \left(\frac{x+y/2+y/2}{3}\right)^3 = \frac{4}{27} (x+y)^3 $$ (look also for weighted AM-GM).

For the second, use one of Maclaurin's inequalities in three variables to get $$ (yz+yw+zw) \leq \frac{1}{3}\left(y+z+w\right)^2, $$ and then the inequality can be reduced to the previous one by setting $Y=y+z+w$.

This kind of tricks can work in simple cases, or where there is much symmetry in the variables (check Muirhead's inequality for instance for another highly-symmetric case); if this is not your case you may be out of luck though.

I do not know how far these methods can go, but in both your examples the maximum can be computed easily and without calculus using some algebraic inequality techniques, the kind of "standard tricks" that is taught to contestants in high-school math Olympiads.

For the first inequality, AM-GM suffices: $$ xy^2 = 4 (x \cdot y/2 \cdot y/2) \leq 4 \left(\frac{x+y/2+y/2}{3}\right)^3 = \frac{4}{27} (x+y)^3 $$ (look also for weighted AM-GM).

For the second, use one of Maclaurin's inequalities in three variables to get $$ (yz+yw+zw) \leq \frac{1}{3}\left(y+z+w\right)^2, $$ and then the inequality can be reduced to the previous one by setting $Y=y+z+w$.

This kind of tricks can work in simple cases, or where there is much symmetry in the variables (check Muirhead's inequality for instance for another highly-symmetric case); if this is not your case you may be out of luck though.

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Federico Poloni
Federico Poloni
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I do not know how far these methods can go, but in both your examples the maximum can be computed easily and without calculus using some algebraic inequality techniques, the kind of "standard tricks" that is taught to contestants in high-school math Olympiads.

For the first inequality, AM-GM suffices: $$ xy^2 = 4 (x \cdot y/2 \cdot y/2) \leq 4 \left(\frac{x+y/2+y/2}{3}\right)^3 = \frac{4}{27} (x+y)^3 $$ (look also for weighted AM-GM).

For the second, use one of Maclaurin's inequalities in three variables to get $$ (yz+yw+zw) \leq \frac{1}{3}\left(y+z+w\right)^2, $$ and then the inequality can be reduced to the previous one by setting $Y=y+z+w$.

This kind of tricks can work in simple cases, or where there is much symmetry in the variables (check Muirhead's inequality for instance for another highly-symmetric case); if this is not your case you may be out of luck though.