- #1

Vriska

- 138

- 2

## Homework Statement

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this is the problem , if x and y are real positive numbers , I need to prove

$$4x^4 + 4y^3 + 5x^2 + y + 1 \ge 12xy$$

## Homework Equations

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$$x^2 + y^2 \ge 2xy$$ (Variation of AM GM Theorem)

## 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.