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\(VT\ge a+b+c+\dfrac{9}{2\left(ab+bc+ca\right)}\ge\sqrt{3\left(ab+bc+ca\right)}+\dfrac{9}{2\left(ab+bc+ca\right)}\)
\(=\dfrac{\sqrt{3\left(ab+bc+ca\right)}}{2}+\dfrac{\sqrt{3\left(ab+bc+ca\right)}}{2}+\dfrac{9}{2\left(ab+bc+ca\right)}\ge3\sqrt[3]{\dfrac{27}{8}}=\dfrac{9}{2}\)
Áp dụng BĐT Cauchy ta có
\(\dfrac{b^2}{a}+a\ge2b;\) \(\dfrac{c^2}{b}+b\ge2c\); \(\dfrac{a^2}{c}+c\ge2a\)
\(\Rightarrow\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}\ge a+b+c\)
\(\Rightarrow\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}+\dfrac{9}{2\left(ab+bc+ac\right)}\ge a+b+c+\dfrac{9}{2\left(ab+bc+ac\right)}\)Ta phải chứng minh
\(a+b+c+\dfrac{9}{2\left(ab+bc+ac\right)}\ge\dfrac{9}{2}\)
\(\Leftrightarrow4\left(a+b+c\right)\left(ab+bc+ac\right)+18\ge18\left(ab+bc+ac\right)\)
\(\Leftrightarrow\left(ab+bc+ac\right)\left(4\left(a+b+c\right)-18\right)+18\ge0\)
Áp dụng BĐT Cauchy:
\(ab+bc+ac\ge3\sqrt[3]{a^2b^2c^2}=3\)
\(a+b+c\ge3\sqrt[3]{abc}=3\)
\(\Rightarrow\left(ab+bc+ac\right)\left(4\left(a+b+c\right)-18\right)+18\ge3\left(4.3-18\right)+18=0\)=> đpcm
Áp dụng BĐT Cauchy Swarch
\(\Sigma\dfrac{1}{a^2+2bc}\ge\dfrac{9}{\left(a+b+c\right)^2}=9\)
Vậy Min ... =9 khi a=b=c=1/3
Ta có \(\sqrt{a-1}+\dfrac{1}{\sqrt{a-1}}\) \(=\sqrt{a-1}+\dfrac{1}{4\sqrt{a-1}}+\dfrac{3}{4\sqrt{a-1}}\) \(\ge2\sqrt{\sqrt{a-1}.\dfrac{1}{4\sqrt{a-1}}}+\dfrac{3}{4\sqrt{a-1}}\) \(=1+\dfrac{3}{4\sqrt{a-1}}\).
Lập 2 BĐT tương tự rồi cộng vế theo vế, ta có
\(VT\ge3+\dfrac{3}{4}\left(\dfrac{1}{\sqrt{a-1}}+\dfrac{1}{\sqrt{b-1}}+\dfrac{1}{\sqrt{c-1}}\right)\)
\(\ge3+\dfrac{3}{4}.\dfrac{9}{\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}}\)
\(\ge3+\dfrac{3}{4}.\dfrac{9}{\dfrac{3}{2}}\) \(=\dfrac{15}{2}\).
ĐTXR \(\Leftrightarrow a=b=c=\dfrac{5}{4}\). Ta có đpcm
Có \(\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}+\dfrac{1}{\sqrt{a-1}}+\dfrac{1}{\sqrt{b-1}}+\dfrac{1}{\sqrt{c-1}}\ge\dfrac{15}{2}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{a-1}}+\dfrac{1}{\sqrt{b-1}}+\dfrac{1}{\sqrt{c-1}}\ge\dfrac{15}{2}-\left(\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\right)\ge6\) (1)
Ta chứng minh (1) đúng
Áp dụng bất đẳng thức Schwarz :
\(\dfrac{1}{\sqrt{a-1}}+\dfrac{1}{\sqrt{b-1}}+\dfrac{1}{\sqrt{c-1}}\ge\dfrac{\left(1+1+1\right)^2}{\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}}\ge\dfrac{9}{\dfrac{3}{2}}=6\)Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\sqrt{a-1}=\sqrt{b-1}=\sqrt{c-1}\\\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}=\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow a=b=c=\dfrac{5}{4}\)(tm)
Áp dụng BĐT Cauchy - Schwarz vào bài toán , ta có :
\(Q=\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ac}+\dfrac{1}{c^2+2ab}\ge\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ac}=\dfrac{9}{\left(a+b+c\right)^2}=\dfrac{9}{1^2}=9\) Dấu " = " xảy ra khi : \(\dfrac{1}{a^2+2ab}=\dfrac{1}{b^2+2ac}=\dfrac{1}{c^2+2ab}\Leftrightarrow a=b=c=\dfrac{1}{3}\)
\(\Rightarrow Q_{Min}=9\Leftrightarrow a=b=c=\dfrac{1}{3}\)
\(1,\text{Áp dụng Mincopxki: }\\ Q\ge\sqrt{\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2}\ge\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}\\ \text{Dấu }"="\Leftrightarrow a=b\)
\(2,\text{Áp dụng BĐT Cauchy-Schwarz: }\\ P\ge\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}=\dfrac{9}{\left(a+b+c\right)^2}\ge\dfrac{9}{1}=9\\ \text{Dấu }"="\Leftrightarrow a=b=c=\dfrac{1}{3}\)
Bài 1:ta có BĐt \(a^3+b^3\ge ab\left(a+b\right)\)vì nó tương đương với \(\left(a+b\right)\left(a-b\right)^2\ge0\)(luôn đúng với a,b>0)
Áp dụng vào bài toán:
\(\dfrac{a^3+b^3}{2ab}+\dfrac{b^3+c^3}{2bc}+\dfrac{c^3+a^3}{2ac}\ge\dfrac{ab\left(a+b\right)}{2ab}+\dfrac{bc\left(b+c\right)}{2bc}+\dfrac{ca\left(c+a\right)}{2ac}=a+b+c\)dấu = xảy ra khi a=b=c
bài 2:
cần chứng minh \(\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}\ge0\)
hay \(\dfrac{a-b}{b+c}+1+\dfrac{b-c}{c+d}+1+\dfrac{c-d}{d+a}+1+\dfrac{d-a}{a+b}+1\ge4\)
\(\Leftrightarrow\dfrac{a+c}{b+c}+\dfrac{b+d}{c+d}+\dfrac{c+a}{d+a}+\dfrac{d+b}{a+b}\ge4\)
xét \(VT=\left(a+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+d}\right)+\left(b+d\right)\left(\dfrac{1}{c+d}+\dfrac{1}{a+b}\right)\)
Áp dụng BĐT cauchy dạng phân thức:
\(\dfrac{1}{b+c}+\dfrac{1}{a+d}\ge\dfrac{4}{a+b+c+d};\dfrac{1}{c+d}+\dfrac{1}{a+b}\ge\dfrac{4}{a+b+c+d}\)
do đó \(VT\ge\dfrac{4\left(a+c\right)}{a+b+c+d}+\dfrac{4\left(b+d\right)}{a+b+c+d}=4\)
dấu = xảy ra khi a=b=c=d
\(a+b+c\le1\) hoặc \(a+b+c=1\) nhá
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT\ge\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=\dfrac{9}{\left(a+b+c\right)^2}=9\)
Đẳng thức xảy ra khi ..........