Cho a,b,c là các số dương. CM BĐT \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)
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12 tháng 11 2017
BĐT cần chứng minh tương đương :
\(\sqrt{\dfrac{a^2+b^2}{2}}-\sqrt{ab}\ge\dfrac{a+b}{2}-\dfrac{2ab}{a+b}\)
\(\Leftrightarrow\dfrac{\dfrac{a^2+b^2}{2}-ab}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\ge\dfrac{\left(a+b\right)^2-4ab}{2\left(a+b\right)}\)
\(\Leftrightarrow\dfrac{\dfrac{\left(a-b\right)^2}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\ge\dfrac{\left(a-b\right)^2}{2\left(a+b\right)}\)
\(\Leftrightarrow\dfrac{\dfrac{\left(a-b\right)^2}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}-\dfrac{\left(a-b\right)^2}{2\left(a+b\right)}\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(\dfrac{\dfrac{1}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}-\dfrac{1}{2\left(a+b\right)}\right)\ge0\)
ta phải chứng minh;
\(\dfrac{\dfrac{1}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}-\dfrac{1}{2\left(a+b\right)}\ge0\)
\(\Leftrightarrow\)\(\dfrac{\dfrac{1}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\ge\dfrac{1}{2\left(a+b\right)}\)
\(\Leftrightarrow a+b\ge\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\)\(\Leftrightarrow2a+2b-\sqrt{2\left(a^2+b^2\right)}-2\sqrt{ab}\ge0\)
\(\Leftrightarrow\left(a+b-\sqrt{2\left(a^2+b^2\right)}\right)+\left(a+b-2\sqrt{ab}\right)\ge0\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2-2\left(a^2+b^2\right)}{a+b+\sqrt{2\left(a^2+b^2\right)}}+\dfrac{\left(a+b\right)^2-4ab}{a+b+2\sqrt{ab}}\ge0\)
\(\Leftrightarrow\dfrac{-\left(a-b\right)^2}{a+b+\sqrt{2\left(a^2+b^2\right)}}+\dfrac{\left(a-b\right)^2}{a+b+2\sqrt{ab}}\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(\dfrac{1}{a+b+2\sqrt{ab}}-\dfrac{1}{a+b+\sqrt{2\left(a^2+b^2\right)}}\right)\ge0\)
ta phải chứng minh
\(\Leftrightarrow\dfrac{1}{a+b+2\sqrt{ab}}-\dfrac{1}{a+b+\sqrt{2\left(a^2+b^2\right)}}\ge0\)
\(\Leftrightarrow\dfrac{1}{a+b+2\sqrt{ab}}\ge\dfrac{1}{a+b+\sqrt{2\left(a^2+b^2\right)}}\)
\(\Leftrightarrow a+b+2\sqrt{ab}\le a+b+\sqrt{2\left(a^2+b^2\right)}\)
\(\Leftrightarrow2\sqrt{ab}\le\sqrt{2\left(a^2+b^2\right)}\Leftrightarrow\left(a-b\right)^2\ge0\)
TM
1
17 tháng 4 2023
\(VT=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)+\left(\dfrac{c}{a+b}+1\right)-3\)
\(=\dfrac{1}{2}\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)-3>=\dfrac{9}{2}-3=\dfrac{3}{2}\)
2 tháng 8 2017
Nice proof, nhưng đã quy đồng là phải thế này :v
\(BDT\Leftrightarrow\left(2a-\sqrt{a^2+3}\right)+\left(2b-\sqrt{b^2+3}\right)+\left(2c-\sqrt{c^2+3}\right)\)
\(\Leftrightarrow\dfrac{a^2-1}{2a+\sqrt{a^2+3}}+\dfrac{b^2-1}{2b+\sqrt{b^2+3}}+\dfrac{c^2-1}{2c+\sqrt{c^2+3}}\ge0\)
\(\Leftrightarrow\dfrac{a^2-1}{2a+\sqrt{a^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{a}-a\right)+\dfrac{b^2-1}{2b+\sqrt{b^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{b}-b\right)+\dfrac{c^2-1}{2c+\sqrt{c^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{c}-c\right)\ge0\)
\(\Leftrightarrow\left(a^2-1\right)\left(\dfrac{1}{2a+\sqrt{a^2+3}}-\dfrac{1}{4a}\right)+\left(b^2-1\right)\left(\dfrac{1}{2b+\sqrt{b^2+3}}-\dfrac{1}{4b}\right)+\left(c^2-1\right)\left(\dfrac{1}{2c+\sqrt{a^2+3}}-\dfrac{1}{4c}\right)\ge0\)
\(\Leftrightarrow\dfrac{\left(a^2-1\right)\left(2a-\sqrt{a^2+3}\right)}{a\left(2a+\sqrt{a^2+3}\right)}+\dfrac{\left(b^2-1\right)\left(2b-\sqrt{b^2+3}\right)}{b\left(2b+\sqrt{b^2+3}\right)}+\dfrac{\left(c^2-1\right)\left(2c-\sqrt{c^2+3}\right)}{c\left(2c+\sqrt{c^2+3}\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(a^2-1\right)^2}{a\left(2a+\sqrt{a^2+3}\right)^2}+\dfrac{\left(b^2-1\right)^2}{b\left(2b+\sqrt{b^2+3}\right)^2}+\dfrac{\left(c^2-1\right)^2}{c\left(2c+\sqrt{c^2+3}\right)^2}\ge0\) (luôn đúng)
2 tháng 8 2017
Khi \(f\left(t\right)=\sqrt{1+t}\) là hàm lõm trên \([-1, +\infty)\) ta có:
\(f(t)\le f(3)+f'(3)(t-3)\forall t\ge -1\)
Tức là \(f\left(t\right)\le2+\dfrac{1}{4}\left(t-3\right)=\dfrac{5}{4}+\dfrac{1}{4}t\forall t\ge-1\)
Áp dụng BĐT này ta có:
\(\sqrt{a^2+3}=a\sqrt{1+\dfrac{3}{a^2}}\le a\left(\dfrac{5}{4}+\dfrac{1}{4}\cdot\dfrac{3}{a^2}\right)=\dfrac{5}{4}a+\dfrac{3}{4}\cdot\dfrac{1}{a}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\sqrt{b^2+3}\le\dfrac{5}{4}b+\dfrac{3}{4}\cdot\dfrac{1}{b};\sqrt{c^2+3}\le\dfrac{5}{4}c+\dfrac{3}{4}\cdot\dfrac{1}{c}\)
Cộng theo vế 3 BĐT trên ta có:
\(VP\le\dfrac{5}{4}\left(a+b+c\right)+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=2\left(a+b+c\right)=VT\)
KK
19 tháng 4 2018
áp dụngBĐt cô si cho 2 số ta có
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{a^2}{b^2}.\dfrac{b^2}{c^2}}=2\dfrac{a}{c}\)
tt ta có
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\dfrac{b}{a}\); \(\dfrac{b^2}{a^2}+\dfrac{a^2}{c^2}\ge2\dfrac{b}{c}\)
cộng các BĐT trên ta có
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
⇔ \(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\) (đpcm)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 12 2020
\(\left(a+b+c\right)\left(\dfrac{a}{\left(b+c\right)^2}+\dfrac{b}{\left(c+a\right)^2}+\dfrac{c}{\left(a+b\right)^2}\right)\ge\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2\ge\dfrac{9}{4}\)
\(\Rightarrow\dfrac{a}{\left(b+c\right)^2}+\dfrac{b}{\left(c+a\right)^2}+\dfrac{c}{\left(a+b\right)^2}\ge\dfrac{9}{4\left(a+b+c\right)}\)
Dấu "=" xảy ra khi \(a=b=c\)
LB
2
13 tháng 10 2021
Áp dụng BĐT Cauchy cho 2 số dương:
\(\left\{{}\begin{matrix}\dfrac{a^2}{b}+b\ge2\sqrt{\dfrac{a^2}{b}.b}=2a\\\dfrac{b^2}{c}+c\ge2\sqrt{\dfrac{b^2}{c}.c}=2b\\\dfrac{c^2}{a}+a\ge2\sqrt{\dfrac{c^2}{a}.a}=2c\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}+a+b+c\ge2a+2b+2c\)
\(\Rightarrow\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+b+c\left(đpcm\right)\)
Dấu "=" xay ra \(\Leftrightarrow a=b=c\)
13 tháng 10 2021
Áp dụng BĐT cosi cho 3 số a,b,c dương:
\(\dfrac{a^2}{b}+b\ge2\sqrt{\dfrac{a^2b}{b}}=2a\\ \dfrac{b^2}{c}+c\ge2\sqrt{\dfrac{b^2c}{c}}=2b\\ \dfrac{c^2}{a}+a\ge2\sqrt{\dfrac{c^2a}{a}}=2c\)
Cộng vế theo vế 3 BĐT trên
\(\Leftrightarrow\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}+a+b+c\ge2\left(a+b+c\right)\\ \Leftrightarrow\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+b+c\)
Dấu \("="\Leftrightarrow a=b=c\)
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