Consider the functional: :$\forall t \in \closedint 0 1: \displaystyle \int_a^b \paren {t \paren {P h^2 + Q h'^2} + \paren {1 - t} h'^2} \rd x$ By assumption: :$\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x \ge 0$ For $t = 1$, Euler's Equation reads: :$\map {h''} x = 0$ which, along with condition $\map h a = 0$, is solved by: :$\map h x = x - a$  for which there are no conjugate points in $\closedint a b$. In other words: :$\forall x \in \openint a b: \map h x > 0$ Hence: :$\forall t \in \closedint 0 1: \displaystyle \int_a^b \paren {t \paren {P h'^2 + Q h^2} + \paren {1 - t} h'^2} \rd x \ge 0$ The corresponding Euler's Equation is: :$2 Q h t - \map {\dfrac \d {\d x} } {2 t P h' + 2 h' \paren {1 - t} } = 0$ which is equivalent to: :$-\map {\dfrac \d {\d x} } {\paren {t P + \paren {1 - t} } h'} + t Q h = 0$ Let $\map h {x, t}$ be a solution to this such that: :$\forall t \in \closedint 0 1: \map h {a, t} = 0$ :$\map {h_x} {a, t} = 1$ {{explain|What is $h_x$?}} Suppose that for $\map h {x, t}$ there exists a conjugate point $\tilde a$ to $a$ in $\closedint a b$. In other words: :$\exists \tilde a \in \closedint a b: \map h {\tilde a, 1} = 0$ By definition, $a \ne \tilde a$. Suppose $\tilde a = b$. Then by lemma 1 of Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite: :$\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x = 0$ This agrees with the assumption. Therefore, it is allowed that $\tilde a = b$. For $t = 1$, any other conjugate point of $\map h {x, t}$ may reside only in $\openint a b$. Consider the following set of all points $\tuple {x, t}$: :$\set {\tuple {x, t}: \paren {\forall x \in \closedint a b} \paren {\forall t \in \closedint 0 1} \paren {\map h {x, t} = 0} }$ If it is non-empty, it represents a curve in $x - t$ plane, such that $h_x \left({x, t}\right) \ne 0$. {{explain|Rather than using "it", give it a name and reference that name.}} By the Implicit Function Theorem, $\map x t$ is continuous. By hypothesis, $\tuple {\tilde a, 1}$ lies on this curve. Suppose that the curve starts at this point. The curve can terminate either inside the rectangle or its boundary. If it terminates inside the rectangle $\closedint a b \times \closedint 0 1$, it implies that there is a discontinuous jump in the value of $h$. {{explain|Again, rather than using "it", refer to the object in question directly, so it is clear what is being referred to.}} :Therefore, it contradicts the continuity of $\map h {x, t}$ in the interval $t \in \closedint 0 1$. {{explain|Specify what contradicts what, by invoking the AimForCont construct}} If it intersects the line segment $x = b, 0 \le t \le 1$, then by lemma 2 of Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite it vanishes. {{explain|"it" again}} :This contradicts positive-definiteness of the functional for all $t$. {{explain|See above note on the above contradiction proof}} If it intersects the line segment $a \le x \le b, t = 1$, then $\exists t_0: \paren {\map h {x, t_0} = 0} \land \paren {\map {h_x} {x, t_0} = 0}$. {{explain|$h_x$ again}} If it intersects $a \le x \le b, t = 0$, then Euler's equation reduces to $h'' = 0$ with solution $h = x - a$, which vanishes only for $x = a$. If it intersects $x = a, 0 \le t \le 1$, then $\exists t_0: \map {h_x} {a, t_0} = 0$ {{explain|"it" again}} {{Help|explain cases $t = 1$ and $x = a$}} By Proof by Cases, no such curve exists. Thus, the point $\tuple {\tilde a, 1}$ does not exist, since it belongs to this curve. Hence there are no conjugate points of $\map h {x, 1} = \map h x$ in the interval $\openint a b$. {{qed}}