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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 \(\Leftrightarrow a=b=c\)
\(A=\dfrac{1}{2a-a^2}+\dfrac{1}{2b-b^2}+\dfrac{1}{2c-c^2}+3\\ =\dfrac{1}{2a-a^2}+\dfrac{1}{2b-b^2}+\dfrac{1}{2c-c^2}+3\\ =\left(\dfrac{1}{2a-a^2}+\dfrac{1}{2b-b^2}+\dfrac{1}{2c-c^2}\right)+3\\ \overset{AM-GM}{\ge}\dfrac{9}{2a-a^2+2b-b^2+2c-c^2}+3\\ =\dfrac{9}{\left(2a+2b+2c\right)-\left(a^2+b^2+c^2\right)}+3\\ =\dfrac{9}{\left(2a+2b+2c\right)-\left(a^2+b^2+c^2\right)}+3\\ \ge\dfrac{9}{2\left(a+b+c\right)-\dfrac{\left(a+b+c\right)^2}{3}}+3\\ =\dfrac{9}{2\cdot1-\dfrac{1}{3}}+3=\dfrac{42}{5}\)
Dấu \("="\) xảy ra khi : \(\left\{{}\begin{matrix}2a-a^2=2b-b^2=2c-c^2\\a=b=c\\a+b+c=1\end{matrix}\right.\Leftrightarrow a=b=c=\dfrac{1}{3}\)
Vậy \(A_{Min}=\dfrac{42}{5}\) khi \(a=b=c=\dfrac{1}{3}\)
Áp dụng bất đẳng thức: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\) \(\Leftrightarrow a^2+2ab+b^2\ge4ab\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow\left(a-b\right)^2\ge0\left(đúng\right)\)
\(\dfrac{1}{2a+b+c}=\dfrac{1}{4}.\dfrac{4}{2a+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{2a}+\dfrac{1}{b+c}\right)\le\dfrac{1}{4}\left[\dfrac{1}{2a}+\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\right]=\dfrac{1}{8}\left(\dfrac{1}{a}+\dfrac{1}{2b}+\dfrac{1}{2c}\right)\)
CMTT \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{a+2b+c}\le\dfrac{1}{8}\left(\dfrac{1}{2a}+\dfrac{1}{b}+\dfrac{1}{2c}\right)\\\dfrac{1}{a+b+2c}\le\dfrac{1}{8}\left(\dfrac{1}{2a}+\dfrac{1}{2b}+\dfrac{1}{c}\right)\end{matrix}\right.\)
\(\Rightarrow M=\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{8}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{2}{2a}+\dfrac{2}{2b}+\dfrac{2}{2c}\right)=\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{4}.4=1\)
\(minM=1\Leftrightarrow a=b=c=\dfrac{3}{4}\)
12. Ta có \(ab\le\frac{a^2+b^2}{2}\)
=> \(a^2-ab+3b^2+1\ge\frac{a^2}{2}+\frac{5}{2}b^2+1\)
Lại có \(\left(\frac{a^2}{2}+\frac{5}{2}b^2+1\right)\left(\frac{1}{2}+\frac{5}{2}+1\right)\ge\left(\frac{a}{2}+\frac{5}{2}b+1\right)^2\)
=> \(\sqrt{a^2-ab+3b^2+1}\ge\frac{a}{4}+\frac{5b}{4}+\frac{1}{2}\)
=> \(\frac{1}{\sqrt{a^2-ab+3b^2+1}}\le\frac{4}{a+b+b+b+b+b+1+1}\le\frac{4}{64}.\left(\frac{1}{a}+\frac{5}{b}+2\right)\)
Khi đó
\(P\le\frac{1}{16}\left(6\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+6\right)\le\frac{3}{2}\)
Dấu bằng xảy ra khi a=b=c=1
Vậy \(MaxP=\frac{3}{2}\)khi a=b=c=1
13. Ta có \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\le1\)
\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{9}{a+b+c+3}\)( BĐT cosi)
=> \(1\ge\frac{9}{a+b+c+3}\)
=> \(a+b+c\ge6\)
Ta có \(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
=> \(\frac{a^3-b^3}{a^2+ab+b^2}=a-b\)
Tương tự \(\frac{b^3-c^3}{b^2+bc+c^2}=b-c\),,\(\frac{c^3-a^2}{c^2+ac+a^2}=c-a\)
Cộng 3 BT trên ta có
\(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+c^2}=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{c^2+bc+b^2}+\frac{a^3}{a^2+ac+c^2}\)
Khi đó \(2P=\frac{a^3+b^3}{a^2+ab+b^2}+...\)
=> \(2P=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}+....\)
Xét \(\frac{a^2-ab+b^2}{a^2+ab+b^2}\ge\frac{1}{3}\)
<=> \(3\left(a^2-ab+b^2\right)\ge a^2+ab+b^2\)
<=> \(a^2+b^2\ge2ab\)(luôn đúng )
=> \(2P\ge\frac{1}{3}\left(a+b+b+c+a+c\right)=\frac{2}{3}.\left(a+b+c\right)\ge4\)
=> \(P\ge2\)
Vậy \(MinP=2\)khi a=b=c=2
Lưu ý : Chỗ .... là tương tự
Dự đoán xảy ra cực trị khi a = b = c =2. Khi đó P =\(\frac{3\sqrt{2}}{4}\). Ta sẽ chứng minh đó là MAX của P
Ta có: \(\left(\frac{a+b+c}{3}\right)^3-\left(a+b+c\right)\ge abc-\left(a+b+c\right)=2\)
Đặt a + b +c = t>0 suy ra \(\frac{t^3-27t}{27}\ge2\Leftrightarrow t^3-27t\ge54\Leftrightarrow t^3-27t-54\ge0\)
\(\Leftrightarrow\orbr{\begin{cases}t\ge6\\t=-3\left(L\right)\end{cases}}\). Do vậy \(t\ge6\) (em làm tắt xiu nhé,dài quá)
\(P=\Sigma_{cyc}\frac{2}{\sqrt{2}.\sqrt{2\left(a^2+b^2\right)}}\le\sqrt{2}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
Giờ đi chứng minh \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\le\frac{3}{4}\)
Em cần suy ra nghĩ tiếp:(