cho a,b,c>0 và a+b+c = 6.tìm max P= ∑ $\frac{ab}{\sqrt{a ²+b ²+2c ²}}$

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1. c1

   \begin{cases} (ab)/(a^2+b^2+2c^2)\\ (ac)/(a^2+2b^2+c^2)\\ (cb)/(2a^2+b^2+c^2)\end{cases}

   ⇔\begin{cases}≤ (ab)/(1/2((a+c)^2+(b+c)^2))\\≤ (ac)/(1/2((a+b)^2+(b+c)^2))\\ ≤(cb)/(1/2((a+c)^2+(b+a)^2))\end{cases}

   ⇔\begin{cases}≤ (ab)/((a+c)(b+c))\\ ≤(ac)/((a+b)(b+c))\\≤ (cb)/((a+c)(b+a))\end{cases}

   `⇒P≤(ab)/((a+c)(b+c))+(ac)/((a+b)(b+c))+(cb)/((a+c)(b+a))`

   `⇒P≤(ab)/((a+c)(b+c))+(ac)/((a+b)(b+c))+(cb)/((a+c)(b+a))` `⇒P≤c^2/(a+c).c^2/(b+c)+b^2/(a+b).b^2/(b+c)+a^2/(a+c).a^2/(b+a)` `⇒P≤1/2 ((c)/(a+c)+(c)/(b+c)+(b)/(a+b)+(b)/(b+c)+(a)/(a+c)+(a)/(b+a))`

   `⇒P≤ (b(a+c))/(a+c)+(a(b+c))/(b+c)+(c(a+b))/(a+b))`

   `⇒P≤(1+1+1)`

   `⇒P≤3`

   `”=”`xẩy ra khi :
   `a=b=c=2`

   vậy` maxP=3 `khi `a=b=c=2`

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2. Đáp án: $P\le 3$

   Giải thích các bước giải:

   Ta có:

   $P=\sum\dfrac{ab}{\sqrt{a^2+b^2+2c^2}}$

   $\to P=\sum\dfrac{ab}{\sqrt{\left(a^2+c^2\right)+\left(b^2+c^2\right)}}$

   $\to P\le\sum\dfrac{ab}{\sqrt{\dfrac12\left(a+c\right)^2+\dfrac12\left(b+c\right)^2}}$

   $\to P\le\sum\dfrac{ab}{\sqrt{\dfrac12\left(\left(a+c\right)^2+\left(b+c\right)^2\right)}}$

   $\to P\le\sum\dfrac{ab}{\sqrt{\dfrac12\cdot 2\left(a+c\right)\left(b+c\right)}}$

   $\to P\le\sum\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}$

   $\to P\le\sum\dfrac{\sqrt{ab}}{\sqrt{a+c}}\cdot \dfrac{\sqrt{ab}}{\sqrt{b+c}}$

   $\to P\le\sum\dfrac12\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)$

   $\to P\le\dfrac12\cdot\sum\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)$

   $\to P\le\dfrac12\left(a+b+c\right)$

   $\to P\le 3$

   Dấu = xảy ra khi $a=b=c=2$

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