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.

Ta có: P= \(2a+3b+\dfrac{1}{a}+\dfrac{4}{b}\) = \(\text{​​}\text{​​}(\dfrac{1}{a}+a)+\left(\dfrac{4}{b}+b\right)+\left(a+2b\right)\)

Ta thấy: \(\text{​​}\text{​​}(\dfrac{1}{a}+a)\ge2\sqrt{\dfrac{1}{a}\cdot a}=2\)

             \(\text{​​}\text{​​}\left(\dfrac{4}{b}+b\right)\ge2\sqrt{\dfrac{4}{b}\cdot b}=4\)

Do đó: P \(\ge2+4+5=11\)

Vậy: P(min)=11  khi:  \(\left\{{}\begin{matrix}\dfrac{1}{a}=a\\\dfrac{4}{b}=b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right..\)

Áp dụng BĐT Bunyakovsky, ta có:

\(a+b+c\le\sqrt{3(a^2+b^2+c^2)}=\sqrt{3.3}=3\)

Áp dụng BĐT Cauchy, ta có:

\(A=\sum{\dfrac{1}{\sqrt{1+8a^3}}}=\sum{\dfrac{1}{\sqrt{(2a+1)(4a^2-2a+1)}}} \\\ge\sum{\dfrac{1}{\dfrac{4a^2+2}{2}}}=\sum{\dfrac{1}{2a^2+1}} \)

Ta cần chứng minh: \(\dfrac{1}{2a^2+1}\ge\dfrac{-4}{9}a+\dfrac{7}{9} \\<=>\dfrac{8a^3-14a^2+4a+2}{9(2a^2+1)}\ge0 \\<=>\dfrac{2(a-1)^2(4a+1)}{9(2a^2+1)}\ge0 (luôn\ đúng\ với\ mọi\ a>0) \\->\sum{\dfrac{1}{2a^2+1}}\ge\dfrac{-4}{9}(a+b+c)+\dfrac{21}{9}\ge\dfrac{-4}{9}.3+\dfrac{21}{9}=1 \\->A\ge1 \)

Đẳng thức xảy ra khi a = b = c = 1.

Vậy GTNN của A là 1 (khi a = b = c = 1).

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 \(\left(a;b;c\right)=\left(1;\dfrac{1}{2};\dfrac{1}{3}\right)\)

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 \(\Leftrightarrow a=b=c\)

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 \(a=b=c=\dfrac{\sqrt{3}}{3}\).

                      Bài làm :

Ta có :

\(\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)

\(\Leftrightarrow\frac{4}{a+b}\le\frac{1}{a}+\frac{1}{b}\)

\(\Leftrightarrow\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\left(1\right)\)

Dấu "=" xảy ra khi : a=b

Chứng minh tương tự như trên ; ta có :

\(\hept{\begin{cases}\frac{1}{b+c}\text{≤}\frac{1}{4}\left(\frac{1}{b}+\frac{1}{c}\right)\left(2\right)\\\frac{1}{c+a}\text{≤}\frac{1}{4}\left(\frac{1}{c}+\frac{1}{a}\right)\left(3\right)\end{cases}}\)

Cộng vế với vế của (1) ; (2) ; (3) ; ta được :

\(A\text{≤}\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\text{=}\frac{3}{2}\)

Dấu "=" xảy ra khi ;

\(\hept{\begin{cases}a=b=c\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\end{cases}}\Leftrightarrow a=b=c=1\)

Vậy Max (A) = 3/2 khi a=b=c=1

quản lí tên kiểu j z

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