Oxidation-Reduction in Acid Solution

I need help balancing out the following equation through either oxidation number or half-reaction method:

MnO4(aq) + H2O2(aq) + H(aq) ---> Mn(aq) + O2(g) + H2O(l)
(Step by step would be nice as I am using this to study. Thanks!)

Divide into half-reactions:
MnO4-(aq) → Mn2+(aq)
H2O2(aq) → O2(g)

Balance elements other than O and H - Mn is balanced
Balance O by adding H2O
MnO4−(aq) → Mn2+(aq) + 4 H2O(l) [add 4 H2O to products]
Balance H by adding H+
MnO4−(aq) + 8 H+(aq) → Mn2+(aq) + 4 H2O(l) [add 8 H+ to reactants]
H2O2(aq) → O2(g) + 2 H+(aq) [add 2 H+ to products]
Balance charge by adding e-
MnO4−(aq) + 8 H+(aq) + 5 e- → Mn2+(aq) + 4 H2O(l) [add 5 e- to reactants]
H2O2(aq) → O2(g) + 2 H+(aq) + 2 e- [add 2 e- to products]

Multiply each half-reaction by an integer to equalize the number of electrons
2{MnO4−(aq) + 8 H+(aq) + 5 e- → Mn2+(aq) + 4 H2O(l)} [multiply by 2 to give 10 e-]
5{H2O2(aq) → O2(g) + 2 H+(aq) + 2 e-} [multiply by 5 to give 10 e-]

Add half-reactions and cancel substances that appear as both reactants and products
2 MnO4−(aq) + 16 H+(aq) + 5 H2O2(aq) → 2 Mn2+(aq) + 8 H2O(l) + 5 O2(g) + 10 H+(aq)

The balanced equation in acidic solution
2 MnO4−(aq) + 6 H+(aq) + 5 H2O2(aq) → 2 Mn2+(aq) + 8 H2O(l) + 5 O2(g)