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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?

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    $\begingroup$ just making sure, are you forgetting any parentheses around $ax^2$ or $by^2$? $\endgroup$
    – C Squared
    Aug 29, 2020 at 13:27
  • $\begingroup$ no, $a$ and $b$ are not squared $\endgroup$
    – 1qwertyyyy
    Aug 29, 2020 at 13:28

3 Answers 3

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

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  • $\begingroup$ Thank you for you answer, but i do not really understand part under „discriminant” $\endgroup$
    – 1qwertyyyy
    Aug 29, 2020 at 13:56
  • $\begingroup$ edited my answer a bit, if you are familiar with positive semidefinite matrices and determinant, you can use them too. $\endgroup$ Aug 29, 2020 at 14:05
  • $\begingroup$ But why do you only investigate case x=y=1? $\endgroup$
    – 1qwertyyyy
    Aug 29, 2020 at 14:10
  • $\begingroup$ 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. $\endgroup$ Aug 29, 2020 at 14:45
  • $\begingroup$ ok, but i do not see that there is proven in first proof that inequality is true $\endgroup$
    – 1qwertyyyy
    Aug 29, 2020 at 14:55
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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 $$a+b\leq2.$$

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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 .

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