1. The problem statement, all variables and given/known data this is the problem , if x and y are real positive numbers , I need to prove $$4x^4 + 4y^3 + 5x^2 + y + 1 \ge 12xy$$ 2. Relevant equations $$x^2 + y^2 \ge 2xy$$ (Variation of AM GM Theorem) 3. The attempt at a solution but $$x^2 + y^2 \ge 2xy $$, so $$6x^2 + 6y^2 \ge 12xy $$, from this if $$4x^4 + 4y^3 + 5x^2 + y + 1 \ge 6x^2 + 6y^2$$ then the given inequality holds let's try this $$44x^4 + 4y^3 \ge 6x^2 + 6y^2$$ $$4x^4 + 4y^3 / (6x^2 + 6y^2) \ge 1 $$ $$ 4x^4 + 4y^3 /(6x^2 + 6y^2) - 1 \ge 0 $$ $$4x^4 + 4y^3 - 6x^2 - 6y^2 /(6x^2 + 6y^2) \ge 0$$ , divide both sides by the numerator ( this step feels wrong) $$1/(6x^2 + 6y^2) \ge 0 $$ since this is true the inequality $$4x^4 + 4y^3 \ge 6x^2 + 6y^2$$ holds , so from this we can say that $$4x^4 + 4y^3 + 5x^2 + y + 1 \ge 4x^4 + 4y^3 $$ $$5x^2 + y +1 \ge 0 $$ , this is obviously true and thus the above inequality holds BUT , the inequality $$4x^4 + 4y^3 \ge 6x^2 + 6y^2$$ doesn't hold for x and y = 1 but it does seem to work for all values of greater than 2 . Where have I gone wrong ? There seems to be a better way to solve the problem because solving this in such a round about way seems weird . A better solution (this proof does not work for numbers from 0 - 2 ) to this would be helpful.