Cho các số thực dương a, b thỏa mãn điều kiện: \(a+b< =1\). Tìm GTNN của biểu thức: \(P=\dfrac{b^2}{a^2b^2+b^2+1}+\dfrac{b}{2a}\)
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Theo đề ra, ta có:
\(a^2+b^2+c^2\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)
\(=a^3+b^3+c^3+a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\)
Theo BĐT Cô-si:
\(\left\{{}\begin{matrix}a^3+ab^2\ge2a^2b\\b^3+bc^2\ge2b^2c\\c^3+ca^2\ge2c^2a\end{matrix}\right.\Rightarrow a^2+b^2+c^2\ge3\left(a^2b+b^2c+c^2a\right)\)
Do vậy \(M\ge14\left(a^2+b^2+c^2\right)+\dfrac{3\left(ab+bc+ac\right)}{a^2+b^2+c^2}\)
Ta đặt \(a^2+b^2+c^2=k\)
Luôn có \(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2=1\)
Vì thế nên \(k\ge\dfrac{1}{3}\)
Khi đấy:
\(M\ge14k+\dfrac{3\left(1-k\right)}{2k}=\dfrac{k}{2}+\dfrac{27k}{2}+\dfrac{3}{2k}-\dfrac{3}{2}\ge\dfrac{1}{3}.\dfrac{1}{2}+2\sqrt{\dfrac{27k}{2}.\dfrac{3}{2k}}-\dfrac{3}{2}=\dfrac{23}{3}\)
\(\Rightarrow Min_M=\dfrac{23}{3}\Leftrightarrow a=b=c=\dfrac{1}{3}\).
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Hi vọng là tìm GTLN:
Không mất tính tổng quát, giả sử b, c cùng phía với 1 \(\Rightarrow\left(b-1\right)\left(c-1\right)\ge0\Leftrightarrow bc\ge b+c-1\).
Áp dụng bất đẳng thức AM - GM ta có:
\(4=a^2+b^2+c^2+abc\ge a^2+2bc+abc\Leftrightarrow2bc+abc\le4-a^2\Leftrightarrow bc\left(a+2\right)\le\left(2-a\right)\left(a+2\right)\Leftrightarrow bc+a\le2\)
\(\Rightarrow a+b+c\le3\).
Áp dụng bất đẳng thức Schwarz ta có:
\(P\le\dfrac{ab}{9}\left(\dfrac{1}{a}+\dfrac{2}{b}\right)+\dfrac{bc}{9}\left(\dfrac{1}{b}+\dfrac{2}{c}\right)+\dfrac{ca}{9}\left(\dfrac{1}{c}+\dfrac{2}{a}\right)=\dfrac{1}{9}.3\left(a+b+c\right)=\dfrac{1}{3}\left(a+b+c\right)\le1\).
Đẳng thức xảy ra khi a = b = c = 1.
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Khúc đầu là: \(\dfrac{1}{a^4+b^2+2b^2}\) hay \(\dfrac{1}{a^4+b^2+2ab^2}\) ??
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Đặt \(\left(a;2b;3c\right)=\left(x;y;z\right)\Rightarrow x+y+z=3\)
\(Q=\dfrac{x+1}{1+y^2}+\dfrac{y+1}{1+z^2}+\dfrac{z+1}{1+x^2}\)
Ta có:
\(\dfrac{x+1}{1+y^2}=x+1-\dfrac{\left(x+1\right)y^2}{1+y^2}\ge x+1-\dfrac{\left(x+1\right)y^2}{2y}=x+1-\dfrac{\left(x+1\right)y}{2}\)
Tương tự:
\(\dfrac{y+1}{1+z^2}\ge y+1-\dfrac{\left(y+1\right)z}{2}\) ; \(\dfrac{z+1}{1+x^2}\ge z+1-\dfrac{\left(z+1\right)x}{2}\)
Cộng vế:
\(Q\ge\dfrac{x+y+z}{2}+3-\dfrac{1}{2}\left(xy+yz+zx\right)\)
\(Q\ge\dfrac{x+y+z}{2}+3-\dfrac{1}{6}\left(x+y+z\right)^2=\dfrac{3}{2}+3-\dfrac{9}{6}=3\)
\(Q_{min}=3\) khi \(x=y=z=1\) hay \(\left(a;b;c\right)=\left(1;\dfrac{1}{2};\dfrac{1}{3}\right)\)
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Dấu BĐT bị ngược, sửa đề: \(\dfrac{1}{a^4+b^4+2ab^4}+\dfrac{1}{a^2+b^4+2a^2b^2}\le\dfrac{1}{2}\).
Đặt \(b^2=x\left(x>0\right)\Rightarrow a+x=2ax\).
Khi đó ta cần chứng minh:
\(\dfrac{1}{a^4+x^2+2ax^2}+\dfrac{1}{a^2+x^4+2a^2x}\le\dfrac{1}{2}\)
Áp dụng BĐT AM-GM:
\(\dfrac{1}{a^4+x^2+2ax^2}+\dfrac{1}{a^2+x^4+2a^2x}\)
\(\le\dfrac{1}{2a^2x+2ax^2}+\dfrac{1}{2ax^2+2a^2x}\)
\(=\dfrac{2}{2ax\left(a+x\right)}\)
\(=\dfrac{1}{ax\left(a+x\right)}\)
\(=\dfrac{1}{2a^2x^2}\)
Ta thấy: \(a+x\ge2\sqrt{ax}\)
\(\Leftrightarrow2ax\ge2\sqrt{ax}\)
\(\Leftrightarrow ax-\sqrt{ax}\ge0\)
\(\Leftrightarrow\sqrt{ax}\left(\sqrt{ax}-1\right)\ge0\)
\(\Leftrightarrow\sqrt{ax}\ge1\)
\(\Rightarrow ax\ge1\)
Khi đó: \(\dfrac{1}{2a^2x^2}\le\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{a^4+x^2+2ax^2}+\dfrac{1}{a^2+x^4+2a^2x}\le\dfrac{1}{2}\)
Hay \(\dfrac{1}{a^4+b^4+2ab^4}+\dfrac{1}{a^2+b^4+2a^2b^2}\le\dfrac{1}{2}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng bđt Schwarz ta có:
\(P=\dfrac{a^4}{2ab+3ac}+\dfrac{b^4}{2cb+3ab}+\dfrac{c^4}{2ac+3bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{5\left(ab+bc+ca\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{5\left(a^2+b^2+c^2\right)}=\dfrac{1}{5}\).
Đẳng thức xảy ra khi và chỉ khi \(a=b=c=\dfrac{\sqrt{3}}{3}\).
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Ta có: \(\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{b}\)
\(\Rightarrow bc+ca=2ca\)
\(P=\dfrac{a+b}{2a-b}+\dfrac{c+b}{2c-b}=\dfrac{ac+bc}{2ca-bc}+\dfrac{ca+ab}{2ca-ab}\)
\(=\dfrac{ca+bc}{ab}+\dfrac{ca+ab}{bc}=\dfrac{c}{b}+\dfrac{c}{a}+\dfrac{a}{b}+\dfrac{a}{c}=\dfrac{c+a}{b}+\dfrac{c}{a}+\dfrac{a}{c}\)
Ta có :
\(\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}\ge\dfrac{4}{a+c}\left(\text{Svácxơ}\right)\)\(\Rightarrow c+a\ge2b\)
Áp dụng bđt cô si cho 2 số dương
\(\dfrac{c}{a}+\dfrac{a}{c}\ge2\sqrt{\dfrac{c}{a}.\dfrac{a}{c}}=2\)
\(\Rightarrow P\ge\dfrac{2b}{b}+2=4\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
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Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(T=\frac{\frac{1}{a^2}}{\frac{1}{b}+\frac{1}{c}}+\frac{\frac{1}{b^2}}{\frac{1}{c}+\frac{1}{a}}+\frac{\frac{1}{c^2}}{\frac{1}{a}+\frac{1}{b}}\geq \frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}{2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}=\frac{1}{2}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\)
\(\geq \frac{1}{2}.3\sqrt[3]{\frac{1}{abc}}=\frac{3}{2}\) (theo BĐT AM-GM)
Vậy $T_{\min}=\frac{3}{2}$.
Giá trị này đạt tại $a=b=c=1$