Both models penalise the inclusion of a non-zero coefficient, using a penalty function. LASSO regression penalises in a way that is proportionate to the absolute magnitude of the coefficient, and ridge regression penalises in a way that is proportionate to the square of the coefficient. Neither model penalises inputs in the likelihood function where the coefficient is set to zero. For inputs into the likelihood function where a coefficient is non-zero, LASSO regression penalises values near zero more heavily, whereas ridge regression penalises values far from zero more heavily. (In your question, you seem to be making the error of thinking that the squared value is always more than the absolute value. That is not true. For input values with magnitude less than one, the absolute value is larger than the square.)
Intuition tells us that ridge regression will tend to outperform LASSO regression in cases where the true non-zero coefficients are close to zero, relative to the noise in the regression. In this case, ridge regression penalises these values less, so it is more likely to estimate non-zero values for these coefficients. LASSO regression penalises these coefficients more, so it is more likely to incorrectly estimate them to be zero. On the basis of this intuition, I would recommend that you compare these models for some data generated from a regression with coefficients that are small relative to the noise in the regression. If you were to conduct a simulation study with cases like this, you should find that ridge regression tends to outperform LASSO in these cases.