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Henri
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By Kodaira's lemma, as $L$I don't think that your assertion is bigtrue; for example, we can write $L=A+E$ where $A$ is ample andLazarsfeld gives an example $E$ effective(PAG, 2. Suppose that $m$ is3.3) of a big and nef divisor on a surface such that $mA$its graded algebra is very amplenot finitely generated, andso that the divisor can't be semiample.

But there are some close results for nef and big divisors, or even for good divisors $mA$(when the Kodaira dimensions equals the numerical dimension) as Mourougane and Russo showed $mE$ are Cartier: for example, thenWilson's theorem asserts that for any nef and big divisor on an irreducible projective variety, there exists $H^0(X,mA) \subset H^0(X,mA+mE)$ so$m_0\in \mathbb N$ together with an effective divisor $N$ such that for all $|mL|$$m\geq m_0$, the linear system $|mD-N|$ has no basepointbase-point. (PAG, 2.3.9)

By Kodaira's lemma, as $L$ is big, we can write $L=A+E$ where $A$ is ample and $E$ effective. Suppose that $m$ is such that $mA$ is very ample, and that $mA$ and $mE$ are Cartier, then $H^0(X,mA) \subset H^0(X,mA+mE)$ so that $|mL|$ has no basepoint.

I don't think that your assertion is true; for example, Lazarsfeld gives an example (PAG, 2.3.3) of a big and nef divisor on a surface such that its graded algebra is not finitely generated, so that the divisor can't be semiample.

But there are some close results for nef and big divisors, or even for good divisors (when the Kodaira dimensions equals the numerical dimension) as Mourougane and Russo showed : for example, Wilson's theorem asserts that for any nef and big divisor on an irreducible projective variety, there exists $m_0\in \mathbb N$ together with an effective divisor $N$ such that for all $m\geq m_0$, the linear system $|mD-N|$ has no base-point. (PAG, 2.3.9)

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Henri
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By Kodaira's lemma, as $L$ is big, we can write $L=A+E$ where $A$ is ample and $E$ effective. Suppose that $m$ is such that $mA$ is very ample, and that $mA$ and $mE$ are Cartier, then $H^0(X,mA) \subset H^0(X,mA+mE)$ so that $|mL|$ has no basepoint.