Đáp án:
Ta có :
$P=\dfrac{a}{\sqrt[]{a^2+ab+bc+ca}}$ $+\dfrac{b}{\sqrt[]{b^2+ab+bc+ca}}+$ $\dfrac{c}{\sqrt[]{c^2+ab+bc+ca}}$ ( vì $ab+bc+ac=1_{}$ )
$=\dfrac{a}{\sqrt[]{(a+b)(a+c)}}+$ $\dfrac{b}{\sqrt[]{(a+b)(b+c)}}+$ $\dfrac{c}{\sqrt[]{(b+c)(a+c)}}$
$\leq$ $\dfrac{2a}{(a+b)+(a+c)}+$ $\dfrac{2b}{{(a+b)+(b+c)}}+$ $\dfrac{2c}{{(a+c)+(b+c)}}$
$=\dfrac{2a}{2a+b+c}+$ $\dfrac{2b}{{2b+a+c}}+$ $\dfrac{2c}{{2c+a+b}}$
$\leq$$\dfrac{a}{2}($ $\dfrac{1}{a+b}+$ $\dfrac{1}{a+c})+$ $\dfrac{b}{2}($$\dfrac{1}{a+b}+$ $\dfrac{1}{b+c})+$ $\dfrac{c}{2}($ $\dfrac{1}{a+c}+$ $\dfrac{1}{b+c})$
$=\dfrac{a}{2(a+b)}+$ $\dfrac{a}{2(a+c)}+$ $\dfrac{b}{2(a+b)}+$ $\dfrac{b}{2(b+c)}+$ $\dfrac{c}{2(a+c)}+$ $\dfrac{c}{2(b+c)}$
$=\dfrac{a+b}{2(a+b)}+$ $\dfrac{b+c}{2(b+c)}+$ $\dfrac{a+c}{2(a+c)}$
$=\dfrac{1}{2}+$ $\dfrac{1}{2}+$ $\dfrac{1}{2}=$ $\dfrac{3}{2}$
Dấu "=" xảy ra khi $a=b=c=\dfrac{\sqrt[]{3}}{3}$
Vậy $Max_{P}=$ $\frac{3}{2}$ đạt đc khi $a=b=c=\dfrac{\sqrt[]{3}}{3}$
Giải thích các bước giải :
Bước 3 :
$\leq$ $\dfrac{2a}{(a+b)+(a+c)}+$ $\dfrac{2b}{{(a+b)+(b+c)}}+$ $\dfrac{2c}{{(a+c)+(b+c)}}$
$=\dfrac{2a}{2a+b+c}+$ $\dfrac{2b}{{2b+a+c}}+$ $\dfrac{2c}{{2c+a+b}}$
Giải thích :
áp dụng BĐT cô - si : $\sqrt[]{ab}$ $\leq$ $\dfrac{a+b}{2}$ ta có :
$\sqrt[]{(a+b)(a+c)}$ $\leq$ $\dfrac{a+b+a+c}{2}=$ $\dfrac{2a+b+c}{2}$
⇒$\dfrac{a}{\sqrt[]{(a+b)(a+c)}}$$\leq$ $\dfrac{a}{\dfrac{2a+b+c}{2}}=$ $\dfrac{2a}{2a+b+c}$
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