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Bài 3:
\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\)
\(\Leftrightarrow x^2y^2\left(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\ge\dfrac{4}{xy}.x^2y^2\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)
Ta có:\(a^2-b=b^2-c\)
\(\Leftrightarrow a^2-b^2=b-c\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=b-c\)
\(\Leftrightarrow a+b=\frac{b-c}{a-b}\)
\(\Leftrightarrow a+b+1=\frac{b-c}{a-b}+1\)
\(\Leftrightarrow a+b+1=\frac{a-c}{a-b}\)
Cmtt ta có:
\(\hept{\begin{cases}b^2-c=c^2-a\Leftrightarrow b+c+1=\frac{b-a}{b-c}\\c^2-a=a^2-b\Leftrightarrow c+a+1=\frac{c-b}{c-a}\end{cases}}\)
\(\Rightarrow\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)=\frac{a-c}{a-b}.\frac{b-c}{b-a}.\frac{c-b}{c-a}=-1\)
Cre:mạng
Do \(\left\{{}\begin{matrix}a\ge0\\b\ge1\\a+b+c=5\end{matrix}\right.\) \(\Rightarrow c\le4\)
\(\Rightarrow2\le c\le4\Rightarrow\left(c-2\right)\left(c-4\right)\le0\Rightarrow c^2\le6c-8\)
\(0\le a\le1< 6\Rightarrow a\left(a-6\right)\le0\Rightarrow a^2\le6a\)
\(1\le b\le2< 5\Rightarrow\left(b-1\right)\left(b-5\right)\le0\Rightarrow b^2\le6b-5\)
Cộng vế:
\(a^2+b^2+c^2\le6\left(a+b+c\right)-13=17\)
\(A_{max}=17\) khi \(\left(a;b;c\right)=\left(0;1;4\right)\)
Ta có BDT luôn đúng \(\left(a-b\right)^2\ge0\) \(\Leftrightarrow a^2+b^2\ge2ab\) \(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\). Do \(a^2+b^2\le2\) nên \(2\left(a^2+b^2\right)\le4\).
Do đó \(\left(a+b\right)^2\le4\) \(\Leftrightarrow-2\le a+b\le2\), suy ra đpcm. ĐTXR \(\Leftrightarrow a=b=1\)
Lời giải:
Đặt biểu thức đã cho là $P$
Do $a+b+c=6$ nên:
$P=\frac{ab}{2a+b}+\frac{bc}{2b+c}+\frac{ca}{2c+a}$
$2P=\frac{2ab}{2a+b}+\frac{2bc}{2b+c}+\frac{2ca}{2c+a}$
$=b-\frac{b^2}{2a+b}+c-\frac{c^2}{2b+c}+a-\frac{a^2}{2c+a}$
$=a+b+c-\left(\frac{b^2}{2a+b}+\frac{c^2}{2b+c}+\frac{a^2}{2c+a}\right)$
Áp dụng BĐT Cauchy-Schwarz:
$\left(\frac{b^2}{2a+b}+\frac{c^2}{2b+c}+\frac{a^2}{2c+a}\right)\geq \frac{(b+c+a)^2}{2a+b+2b+c+2c+a}=\frac{a+b+c}{3}$
Do đó: $2P\leq a+b+c-\frac{a+b+c}{3}=\frac{2}{3}(a+b+c)=\frac{2}{3}.6=4$
$\Rightarrow P\leq 2$ (đpcm)
Dấu "=" xảy ra khi $a=b=c=2$
Lời giải:
Đặt biểu thức đã cho là $P$
Do $a+b+c=6$ nên:
$P=\frac{ab}{2a+b}+\frac{bc}{2b+c}+\frac{ca}{2c+a}$
$2P=\frac{2ab}{2a+b}+\frac{2bc}{2b+c}+\frac{2ca}{2c+a}$
$=b-\frac{b^2}{2a+b}+c-\frac{c^2}{2b+c}+a-\frac{a^2}{2c+a}$
$=a+b+c-\left(\frac{b^2}{2a+b}+\frac{c^2}{2b+c}+\frac{a^2}{2c+a}\right)$
Áp dụng BĐT Cauchy-Schwarz:
$\left(\frac{b^2}{2a+b}+\frac{c^2}{2b+c}+\frac{a^2}{2c+a}\right)\geq \frac{(b+c+a)^2}{2a+b+2b+c+2c+a}=\frac{a+b+c}{3}$
Do đó: $2P\leq a+b+c-\frac{a+b+c}{3}=\frac{2}{3}(a+b+c)=\frac{2}{3}.6=4$
$\Rightarrow P\leq 2$ (đpcm)
Dấu "=" xảy ra khi $a=b=c=2$