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\(a^2+b^2-ab\ge\dfrac{1}{2}\left(a+b\right)^2-\dfrac{1}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\dfrac{1}{\sqrt{a^2-ab+b^2}}\le\dfrac{1}{\sqrt{\dfrac{1}{4}\left(a+b\right)^2}}=\dfrac{2}{a+b}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tương tự:
\(\dfrac{1}{\sqrt{b^2-bc+c^2}}\le\dfrac{1}{2}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{c^2-ca+a^2}}\le\dfrac{1}{2}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\)
Cộng vế:
\(P\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
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
Áp dụng BĐT AM-GM: \(VT\le\sum\dfrac{1}{\sqrt{a^2+1}.\sqrt{2a}.2\sqrt{bc}}=\sum\dfrac{1}{2\sqrt{2}\sqrt{a^2+1}}\)
Ta đi chứng minh \(\dfrac{1}{\sqrt{a^2+1}}+\dfrac{1}{\sqrt{b^2+1}}+\dfrac{1}{\sqrt{c^2+1}}\le\dfrac{3}{\sqrt{2}}\)
Giả sử c=max{a, b, c}.Suy ra \(c\ge1\) nên \(ab\le1\). Ta có bổ đề:
\(\dfrac{1}{\sqrt{a^2+1}}+\dfrac{1}{\sqrt{b^2+1}}\le\dfrac{2}{\sqrt{1+ab}}\)(*)
#cm: Áp dụng Bunyakovsky: \(VT_{(*)} \)\(\le\sqrt{2\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\right)}\)
Xét \(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}-\dfrac{2}{ab+1}=\dfrac{\left(a-b\right)^2\left(ab-1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\le0\)
Nên \(VT_{(*)}\)\(\le\sqrt{2.\dfrac{2}{ab+1}}=\dfrac{2}{\sqrt{ab+1}}\), suy ra đpcm.
Do đó \(VT\le\dfrac{2}{\sqrt{ab+1}}+\dfrac{1}{\sqrt{c^2+1}}=2\sqrt{\dfrac{c}{c+1}}+\dfrac{1}{\sqrt{c^2+1}}\)
# cm: \(2\sqrt{\dfrac{c}{c+1}}+\dfrac{1}{\sqrt{c^2+1}}\le\dfrac{3}{\sqrt{2}}\)
\(\Leftrightarrow2\sqrt{2c\left(c^2+1\right)}+\sqrt{2c+2}\le3\sqrt{\left(c+1\right)\left(c^2+1\right)}\)
\(\Leftrightarrow8c^3+10c+2+8\sqrt{c\left(c+1\right)\left(c^2+1\right)}\le9\left(c^3+c^2+c+1\right)\)
hay \(8\sqrt{\left(c^2+c\right)\left(c^2+1\right)}\le c^3+9c^2-c+7\) ($)
Áp dụng BĐT AM-GM cho VT của ($):
\(8\sqrt{\left(c^2+c\right)\left(c^2+1\right)}\le4\left(2c^2+c+1\right)\) .Ta chứng minh
\(8c^2+4c+4\le c^3+9c^2-c+7\) hay \(\left(c-1\right)^2\left(c+3\right)\ge0\) (đúng)
Vậy ta có đpcm. Dấu = xảy ra khi a=b=c=1
Ta có:\(\dfrac{1}{a+b+1}+\dfrac{1}{b+c+1}+\dfrac{1}{a+c+1}=2\)
\(\Rightarrow\dfrac{1}{a+b+1}=\left(1-\dfrac{1}{b+c+1}\right)+\left(1-\dfrac{1}{a+c+1}\right)\)
\(\Rightarrow\dfrac{1}{a+b+1}=\dfrac{b+c}{b+c+1}+\dfrac{a+c}{a+c+1}\ge2\sqrt{\dfrac{\left(b+c\right)\left(a+c\right)}{\left(b+c+1\right)\left(a+c+1\right)}}\)Chứng minh tương tự :\(\dfrac{1}{b+c+1}\ge2\sqrt{\dfrac{\left(a+b\right)\left(a+c\right)}{\left(a+b+1\right)\left(a+c+1\right)}}\)
\(\dfrac{1}{a+c+1}\ge2\sqrt{\dfrac{\left(a+b\right)\left(b+c\right)}{\left(a+b+1\right)\left(b+c+1\right)}}\)
Nhân các bất đẳng thức trên lại với nhau về theo vế ,ta được:
\(\dfrac{1}{\left(a+b+1\right)\left(b+c+1\right)\left(a+c+1\right)}\ge\dfrac{8\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b+1\right)\left(b+c+1\right)\left(a+c+1\right)}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\dfrac{1}{8}\)
Dấu "=" xảy ra khi:\(a=b=c=\dfrac{1}{4}\)
Vậy giá trị lớn nhất của (a+b)(b+c)(c+a) là \(\dfrac{1}{8}\) khi \(a=b=c=\dfrac{1}{4}\)
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 \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{a+2b+c}\le\dfrac{1}{8}\left(\dfrac{1}{2a}+\dfrac{1}{b}+\dfrac{1}{2c}\right)\\\dfrac{1}{a+b+2c}\le\dfrac{1}{8}\left(\dfrac{1}{2a}+\dfrac{1}{2b}+\dfrac{1}{c}\right)\end{matrix}\right.\)
\(\Rightarrow M=\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{8}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{2}{2a}+\dfrac{2}{2b}+\dfrac{2}{2c}\right)=\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{4}.4=1\)
\(minM=1\Leftrightarrow a=b=c=\dfrac{3}{4}\)
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:(
Đặt \(a+b=x,b+c=y,c+a=z\) với \(x,y,z>0\). Ta có:
\(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}=2\)
\(\Rightarrow\dfrac{1}{x+1}=2-\dfrac{1}{y+1}-\dfrac{1}{z+1}\) \(=1-\dfrac{1}{y+1}+1-\dfrac{1}{z+1}\) \(=\dfrac{y}{y+1}+\dfrac{z}{z+1}\)
\(\Rightarrow\dfrac{1}{x+1}\ge2\sqrt{\dfrac{y}{y+1}.\dfrac{z}{z+1}}\)
Tương tự, ta có: \(\dfrac{1}{y+1}\ge2\sqrt{\dfrac{z}{z+1}.\dfrac{x}{x+1}}\) và \(\dfrac{1}{z+1}\ge2\sqrt{\dfrac{x}{x+1}.\dfrac{y}{y+1}}\)
Nhân theo vế 3 BĐT vừa tìm được, ta có:
\(\dfrac{1}{x+1}.\dfrac{1}{y+1}.\dfrac{1}{z+1}\ge2\sqrt{\dfrac{y}{y+1}.\dfrac{z}{z+1}}.2\sqrt{\dfrac{z}{z+1}.\dfrac{x}{x+1}}.2\sqrt{\dfrac{x}{x+1}.\dfrac{y}{y+1}}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge8.\dfrac{xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(\Leftrightarrow xyz\le\dfrac{1}{8}\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\dfrac{1}{8}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{4}\)
Vậy GTLN của \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\) là \(\dfrac{1}{8}\), xảy ra khi \(a=b=c=\dfrac{1}{4}\)