Let $x>1,y>1$ and $\left(\log_e x \right)^2+\left(\log_e y \right)^2=\log(x^2)+log(y^2)$, then find the maximum value of $x^{\log_e y}$

My attempt is as follows:

As $x>1,y>1$ , so $\log_e x>0, \log_e y>0$, hence we can apply $AM>=GM$

$$\dfrac{log_e x+log_e y}{2}>=\sqrt{\log_e x\log_e y}$$

As both the sides are positive, so we can square both the sides without breaking the inequality.

$$(\log_e x)^2+(\log_e y)^2+2\log_e x\log_e y>=4\log_e x\log_e y$$

Using the given condition $\left(\log_e x \right)^2+\left(\log_e y \right)^2=\log(x^2)+log(y^2)$

$$(\log_e x^2)+(\log_e y^2)>=2\log_e x\log_e y$$ $$(\log_e x)+(\log_e y)>=\log_e x\log_e y$$ $$\log_e xy>=\log_e x^{\log_e y}$$

As $e>1$, so we can safely write $xy>=x^{\log_e y}$

But actual answer is $e^4$, I am not able to think of any other way. Please help me in this.

  • $\begingroup$ What do you mean by $\log$? Is this with respect to the root $10$? $\endgroup$ – Allawonder Oct 27 '19 at 5:53

Let $\ln{x}=a$ and $\ln{y}=b$.

Thus, $a$ and $b$ are positives, $$a^2+b^2=2(a+b)$$ and by C-S $$2(a+b)=\frac{1}{2}(1+1)(a^2+b^2)\geq\frac{1}{2}(a+b)^2.$$ Thus, $$a+b\leq4$$ and by AM-GM: $$2\sqrt{ab}\leq a+b\leq4,$$ which gives $$ab\leq4.$$ Id est, $$x^{\ln{y}}=e^{\ln{x}\ln{y}}=e^{ab}\leq e^4.$$ The equality occurs for $x=y=e^2,$ which says that we got a maximal value.

C-S it's the following.

For any $a_i$ and $b_i$ we have: $$(a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2)\geq(a_1b_1+a_2b_2+...+a_nb_n)^2.$$

In our case $n=2$, $a_1=a_2=1$, $b_1=a$ and $b_2=b$.

Thus, $$2(a^2+b^2)=(1^2+1^2)(a^2+b^2)\geq(1\cdot a+1\cdot b)^2=(a+b)^2.$$ By the way, also, you can get the last inequality by the following way: $$2(a^2+b^2)-(a+b)^2=(a-b)^2\geq0.$$

  • $\begingroup$ can you tell me the step titled C-S, I didn't understand $\endgroup$ – user3290550 Oct 27 '19 at 1:58
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    $\begingroup$ It's Cauchy-Schwarz inequality. $\endgroup$ – Michael Rozenberg Oct 27 '19 at 1:59
  • $\begingroup$ can you explain about it a bit here in the current context $\endgroup$ – user3290550 Oct 27 '19 at 2:00
  • $\begingroup$ ok so you are saying that $2(a^2+b^2)>=(a+b)^2$, so how you wrote $a+b<=4$ $\endgroup$ – user3290550 Oct 27 '19 at 2:20
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    $\begingroup$ ok got it , sorry , very nice solution , I forgot that $a^2+b^2=2(a+b)$ in this case $\endgroup$ – user3290550 Oct 27 '19 at 2:23

Here is a way where you do not need any standard inequality like AM-GM or Cauchy-Schwarz.

The fist step is, as already shown in an other solution, to get rid of the logarithms by substituting $x=e^a$, $y=e^b$ with $a,b >0$ which gives:

  • $a^2+b^2 = 2(a+b)$ and
  • maximize $x^{\ln y} =e^{\ln x\cdot \ln y} = e^{ab}$.

Now, observe that by using square completion you get $$a^2+b^2 = 2(a+b) \Leftrightarrow a^2-2a + b^2+2b = 0 \Leftrightarrow \boxed{(a-1)^2+(b-1)^2=2}$$

This is a circle around $(1,1)$ with radius $\sqrt{2}$. So, you can write

  • $a-1 =\sqrt{2}\sqrt{\cos t} \Leftrightarrow \boxed{a = 1+ \sqrt{2}\cos t}$
  • $b-1 =\sqrt{2}\sqrt{\sin t} \Leftrightarrow \boxed{b = 1+ \sqrt{2}\sin t}$

Now, lets maximize $ab$ and let's check at the end whether the maximum is attained for a $t \in [0,2\pi]$ such that $a,b >0$. Here you may use

  • $(1)$: $\sin\frac{\pi}{4} =\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$
  • $(2)$: $(1)\Rightarrow \frac{\sqrt{2}}{2}(\cos t + \sin t) = \sin \left(t+\frac{\pi}{4}\right)$
  • $(3)$: $2\sin t \cos t = \sin 2t$

\begin{eqnarray*}ab & = & (1+ \sqrt{2}\cos t)(1+ \sqrt{2}\sin t)\\ & = & 1+ \sqrt{2}(\cos t + \sin t) + 2\cos t \sin t \\ & \stackrel{(2),(3)}{=} & 1 + 2\sin \left(t+\frac{\pi}{4}\right) + \sin 2t \\ & \leq & 1+2+1 = \boxed{4} \end{eqnarray*} Since the maximum is indeed attained for $\boxed{t=\frac{\pi}{4}}$ which satisfies the condition $a,b>0$, you are done.

So, the maximum value of $x^{\log_e y}$ under the given constraints is $\boxed{e^4}$.


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