# Proving that inequality holds under condition.

Let $$a$$ and $$b$$ be positive numbers. Prove that inequality $$\frac{ax+by}{2} \leqslant \sqrt{\frac{ax^2+by^2}{2}}$$ holds for all real $$x$$ and $$y$$ only and only if $$a+b \leqslant2$$

Problem needs to be done using "basic" algebraic methods.

I tried expanding this into form $$2ax^2+2by^2-a^2x^2-b^2y^2-2abxy \geqslant 0$$ and then take oustide parenthesis $$2-a-b$$. Inequalities between means did not help either. Can you give me some clues?

• just making sure, are you forgetting any parentheses around $ax^2$ or $by^2$? Aug 29, 2020 at 13:27
• no, $a$ and $b$ are not squared Aug 29, 2020 at 13:28

Suppose $$\frac{ax+by}2 \le \sqrt{\frac{ax^2+by^2}{2}}$$ is true, let $$x=y=1$$,

$$\frac{a+b}2\le \sqrt{\frac{a+b}2}$$

$$\frac{(a+b)^2}{4}\le \frac{a+b}2$$

Hence we must have $$a+b \le 2$$.

Suppose we have $$a+b \le 2$$, we want to investigate when does

$$(2a-a^2)x^2+(2b-b^2)y^2-2abxy \ge 0, \forall x, y$$

View it as a quadratic equation in $$x$$, since the coefficient $$2a-a^2$$ is positive, this is equivalent to the discriminant being non-positive. $$4a^2b^2y^2 -4(2a-a^2)(2b-b^2)y^2 \le 0, \forall y$$

Equivalently,

$$ab - (2-a)(2-b) \le 0$$

$$-4+2a+2b \le 0$$

$$a+b \le 2$$

which is true as that is our assumption. That is $$a+b \le 2 \implies \frac{ax+by}2 \le \sqrt{\frac{ax^2+by^2}{2}}$$.

Conclusion: $$a+b \le 2 \iff \frac{ax+by}2 \le \sqrt{\frac{ax^2+by^2}{2}}$$.

• Thank you for you answer, but i do not really understand part under „discriminant” Aug 29, 2020 at 13:56
• edited my answer a bit, if you are familiar with positive semidefinite matrices and determinant, you can use them too. Aug 29, 2020 at 14:05
• But why do you only investigate case x=y=1? Aug 29, 2020 at 14:10
• oh, i tried to work out two direction separately, first I want to show that if the inequality holds, we must have $a+b \le 2$ (which can be proven by taking $x=y=1$). After that I assume $a+ b \le 2$ and show that it is equivalent to the inequality. Aug 29, 2020 at 14:45
• ok, but i do not see that there is proven in first proof that inequality is true Aug 29, 2020 at 14:55

Let $$a+b\leq2$$.

Thus, by C-S $$\sqrt{\frac{ax^2+by^2}{2}}=\sqrt{\frac{(a+b)(ax^2+by^2)}{2(a+b)}}\geq\frac{|ax+by|}{\sqrt{2(a+b)}}\geq\frac{|ax+by|}{2}\geq\frac{ax+by}{2}.$$ Let $$a$$ and $$b$$ are positives and $$\sqrt{\frac{ax^2+by^2}{2}}\geq\frac{ax+by}{2}$$ is true for any reals $$x$$ and $$y$$.

Thus, for $$x=y=1$$ we obtain: $$\sqrt{\frac{a+b}{2}}\geq\frac{a+b}{2},$$ which gives $$a+b\leq2.$$

The inequality is equivalent to :

$$\Big(\frac{ax+by}{2}\Big)^2 \leqslant \frac{ax^2+by^2}{2}$$

By Jensen's inequality and the convexity of $$f(x)=x^2$$ we get :

$$\frac{ax^2+by^2}{2}\geq \Big(\frac{a+b}{2}\Big)\Big(\frac{ax+by}{a+b}\Big)^2$$

Or : $$\frac{ax^2+by^2}{2}\geq \Big(\frac{1}{2(a+b)}\Big)\Big(ax+by\Big)^2$$

But :$$a+b\leq 2$$

Or : $$\frac{1}{a+b}\geq \frac{1}{2}$$

Now you can conclude I think .