Your Approach Should be Optimum

$NOTE:\quad we\quad can\quad write\longrightarrow \quad x+1=\quad (x+\sqrt { 3 } \sqrt { \frac { 1 }{ 3 } } )\\ \{ Now\quad use\quad Cauchy\quad schz.\quad inequality\} \\ \\ \Longrightarrow { \{ x+\sqrt { 3 } \sqrt { \frac { 1 }{ 3 } } \} }^{ 2 }\quad \le \quad { \{ x }^{ 2 }+{ (\sqrt { 3 } })^{ 2 }\} \times \{ { 1 }^{ 2 }+{ (\sqrt { \frac { 1 }{ 3 } } })^{ 2 }\} \\ \\ \Longrightarrow { f\left( x \right) }_{ max }=\frac { 4 }{ 3 } \\ \Longrightarrow a=4\quad ,\quad b=3\\ \Longrightarrow a+b=7\quad$ .

I Think This is best use of Cauchy inequality.

$\frac { \left( { x }+1 \right) ^{ 2 } }{ ({ x }^{ 2 }+3) } \quad =\quad 1\quad +\quad 2\frac { x-1 }{ { x }^{ 2 }+3 } \\ \\ Now\quad maxima\quad of\quad 2\frac { x-1 }{ { x }^{ 2 }+3 } \\ \\ occurs\quad when\quad its\quad derivative\quad is\quad 0\\ \\ which\quad occurs\quad when\quad 2x(x-1)\quad =\quad { x }^{ 2 }+3\\ or\quad at\quad x\quad =\quad -1\quad and\quad at\quad x\quad =3\\ \\ clearly\quad maxima\quad occurs\quad at\quad x\quad =3,\quad \\ hence\quad maximum\quad value\quad is\quad \\ \\ \frac { 4 }{ 3 } \\ \\ or\quad a+b\quad =\quad 7$ .