... curve¹

There is a particular case that should not be ignored: $\nabla h(P)=0$. If the gradient is zero, we can not really say it is perpendicular to some curve. The method of Lagrangian multipliers should check all the points where $\nabla h\rvert_{P}=0$. This is described in greater detail in [“Lagrange Multipliers Can Fail to Determine Extrema,” College Mathematics Journal, Vol. 34, No. 1 (2003), pp. 60–62], see https://www.maa.org/sites/default/files/nunemacher01010325718.pdf.

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....²

If we move away from $\nabla g_1\rvert_{P}$ to some other direction $d$, the hyperplane perpendicular to $d$ in $P$ will not be tangent to the level surface $g_1(x_1,x_2,\dots x_n)=0$. As such, there are curves $r_1$ on the level surface along which $d\cdot r_1'(0)$ can either increase or decrease as we move in either direction from $t=0$.

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... form:³

Care should be taken that there might be other functions $f$ for which $P$ is a constrained maximum. For instance, if $\nabla g_1\rvert_{P}$ and $\nabla h\rvert_{P}$ are linearly dependent, the associated surfaces have the same supporting (tangent) hyperplane in $P$ and the feasible area could be reduced to one point. In such a case, $P$ is the constrained optimum for any function $f$. For instance, if $g_1(x_1,x_2)=x^2_1+x^2_2-1\leq 0$ and $h(x_1,x_2)=x_1-1 = 0$, the only feasible point is $P=(1,0)$. To ensure the necessity of the KKT conditions, one has to assume some regularity conditions.

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