Cho 3 số a,b,c > 0 thỏa mãn a + b + c = 3
Chứng minh : \(\dfrac{a^2}{a+2b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+a^2}\ge1\)
Cho 3 số thực dương a, b, c thỏa mãn: abc=1. Chứng minh rằng:
\(\dfrac{a^3}{a^2+2b^2}+\dfrac{b^3}{b^2+2c^2}+\dfrac{c^3}{c^2+2a^2}\ge1\)
Ta có: \(\dfrac{a^3}{a^2+2b^2}=a-\dfrac{2ab^2}{a^2+2b^2}\ge a-\dfrac{2ab^2}{3\sqrt[3]{a^2b^4}}=a-\dfrac{2}{3}\sqrt[3]{ab^2}\ge a-\dfrac{2}{9}\left(a+b+b\right)=a-\dfrac{2}{9}\left(a+2b\right)\) Chứng minh tương tự ta được:
\(\dfrac{b^3}{b^2+2c^2}\ge b-\dfrac{2}{9}\left(b+2c\right);\dfrac{c^3}{c^2+2a^2}\ge c-\dfrac{2}{9}\left(c+2a\right)\)
\(\Rightarrow\dfrac{a^3}{a^2+2b^2}+\dfrac{b^3}{b^2+2c^2}+\dfrac{c^3}{c^2+2a^2}\ge a+b+c-\dfrac{2}{9}\left(a+2b+b+2c+c+2a\right)=a+b+c-\dfrac{2}{9}\left(3a+3b+3c\right)=\dfrac{1}{3}\left(a+b+c\right)\ge\dfrac{1}{3}\cdot3\sqrt[3]{abc}=1\)Dấu = xảy ra \(\Leftrightarrow a=b=c=1\)
Cho a,b,c là các số thực dương thỏa a+b+c=3
Chứng minh \(\dfrac{1}{2+a^2b}+\dfrac{1}{2+b^2c}+\dfrac{1}{2+c^2a}\ge1\)
Ta có : \(a+b+c=3\Rightarrow a^2+b^2+c^2\ge3\)
Theo BĐT AM - GM ta có :
\(a^4+b^2\ge2a^2b\)
\(b^4+c^2\ge2b^2c\)
\(c^4+a^2\ge2c^2a\)
\(2a^2b^2+2a^2\ge4a^2b\)
\(2b^2c^2+2b^2\ge4b^2c\)
\(2c^2a^2+2c^2\ge4c^2a\)
Cộng từng vế BĐT ta được :
\(\left(a^2+b^2+c^2\right)^2+3\left(a^2+b^2+c^2\right)\ge6\left(a^2b+b^2c+c^2a\right)\)
\(\Rightarrow a^2b+b^2c+c^2a\le\dfrac{3^2+3^2}{6}=3\)
Theo BĐT Cauchy schwarz dưới dạng en-gel ta có :
\(VT\ge\dfrac{9}{6+a^2b+b^2c+c^2a}=\dfrac{9}{9}=1\)
Dấu bằng xảy ra khi \(a=b=c=1\)
Viết lại BĐT:\(\dfrac{a^2b}{a^2b+2}+\dfrac{b^2c}{b^2c+2}+\dfrac{c^2a}{c^2a+2}\le1\)
Áp dụng BĐT AM-GM:
\(VT\le\sum\dfrac{a^2b}{3\sqrt[3]{a^4b^2}}=\dfrac{1}{3}\left(\sqrt[3]{a^2b}+\sqrt[3]{b^2c}+\sqrt[3]{c^2a}\right)\)
\(\le\dfrac{1}{9}\left(3a+3b+3c\right)=1\)
Suy ra đpcm
Cho a,b,c>0 thỏa mãn: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1980\)
Chứng minh rằng: \(\dfrac{\sqrt{b^2+2a^2}}{ab}+\dfrac{\sqrt{c^2+2b^2}}{bc}+\dfrac{\sqrt{a^2+2c^2}}{ac}\ge1980\sqrt{3}\)
\(\dfrac{\sqrt{b^2+a^2+a^2}}{ab}\ge\dfrac{\sqrt{\dfrac{1}{3}\left(b+a+a\right)^2}}{ab}=\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{a}+\dfrac{2}{b}\right)\)
Tương tự: \(\dfrac{\sqrt{c^2+2b^2}}{bc}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{b}+\dfrac{2}{c}\right)\) ; \(\dfrac{\sqrt{a^2+2c^2}}{ac}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{c}+\dfrac{2}{a}\right)\)
Cộng vế với vế:
\(VT\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\sqrt{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1980\sqrt{3}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{3}{1980}\)
Cho a; b; c là các số thực dương thỏa mãn a^2 + b^2 + c^2 = 3
Chứng minh: \(\dfrac{2a^2}{a+b^2}+\dfrac{2b^2}{b+c^2}+\dfrac{2c^2}{c+a^2}\ge a+b+c\)
Ta có \(a+b^2\le\dfrac{a^2+1}{2}+b^2=\dfrac{a^2+2b^2+1}{2}\)
\(\Rightarrow\dfrac{2a^2}{a+b^2}\ge\dfrac{4a^2}{a^2+2b^2+1}=\dfrac{4a^4}{a^4+2b^2a^2+a^2}\). Lập 2 BĐT tương tự rồi áp dụng bất đẳng thức BCS, ta có:
\(\dfrac{2a^2}{a+b^2}+\dfrac{2b^2}{b+c^2}+\dfrac{2c^2}{c+a^2}\ge\dfrac{\left(2a^2+2b^2+2c^2\right)^2}{a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)+a^2+b^2+c^2}\) \(=\dfrac{4\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)^2+3}\)\(=\dfrac{4.3^2}{3^2+3}=3\).
Mà \(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\) nên ta có đpcm. ĐTXR \(\Leftrightarrow a=b=c=1\)
Bài 1: Cho a,b,c là những số dương thỏa mãn: a+b+c=3
CMR: \(\dfrac{a^2}{a+2b^3}+\dfrac{b^2}{b+2c^3}+\dfrac{c^2}{c+2a^3}\ge1\)
Bài 2: Cho a, b, c thỏa mãn: ab+bc+ca=3
CMR: \(\dfrac{a}{2b^3+1}+\dfrac{b}{2c^3+1}+\dfrac{c}{2a^3+1}\ge1\)
Bài 3: Cho a, b, c > 0. CMR: \(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+3b\)
Dấu = xảy ra khi a=b=2c
Bài 5: cho a,b,c lớn hơn 0
chứng minh rẳng:
\(2\left(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}\right)\ge1+\dfrac{b}{b+2a}+\dfrac{c}{c+2b}+\dfrac{a}{a+2c}\)
\(2\left(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}\right)\ge1+\dfrac{b}{b+1a}+\dfrac{c}{c+2b}+\dfrac{a}{a+2c}\)
\(\Leftrightarrow2\left(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}+\dfrac{a}{b+2a}+\dfrac{b}{c+2b}+\dfrac{c}{a+2c}\right)\ge1+\dfrac{b+2a}{b+2a}+\dfrac{c+2b}{c+2b}+\dfrac{a+2c}{a+2c}=1+1+1+1=4\)Thật vậy:
\(\dfrac{a}{b+2c}+\dfrac{a}{b+2a}+\dfrac{b}{c+2a}+\dfrac{b}{c+2b}+\dfrac{c}{a+2b}+\dfrac{c}{a+2c}=a\left(\dfrac{1}{b+2c}+\dfrac{1}{b+2a}\right)+b\left(\dfrac{1}{c+2a}+\dfrac{1}{c+2b}\right)+c\left(\dfrac{1}{a+2b}+\dfrac{1}{a+2c}\right)\)
\(\ge\dfrac{4a}{2\left(a+b+c\right)}+\dfrac{4b}{2\left(a+b+c\right)}+\dfrac{4c}{2\left(a+b+c\right)}=2\)
\(\Rightarrow VT\ge2.2=4\)
\(\RightarrowĐPCM\)
Cho các số thực dương a, b, c thỏa mãn a + b + c = 3. Chứng minh rằng: \(\sqrt{2a^2+\dfrac{7}{b^2}}+\sqrt{2b^2+\dfrac{7}{c^2}}+\sqrt{2c^2+\dfrac{7}{a^2}}\ge9\)
Cho ba số thực dương a,b,c thỏa mãn abc = 1
Chứng minh rằng : \(\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\) ≤ \(\dfrac{1}{2}\)
\(Áp\ dụng\ BĐT\ AM - GM,\ ta\ có: \\\sum\dfrac{1}{a^2+2b^2+3}=\sum\dfrac{1}{(a^2+b^2)+(b^2+1)+2}\le\sum\dfrac{1}{2ab+2b+2} \\=\dfrac{1}{2}\sum\dfrac{1}{ab+b+1}=\dfrac{1}{2}.1=\dfrac{1}{2} \\Đẳng\ thức\ xảy\ ra\ khi\ a=b=c=1.\)
Với a, b, c là những số thực dương thỏa mãn \(\left(a+b\right)\left(b+c\right)\)\(\left(c+a\right)\)=1
Chứng minh rằng \(\dfrac{a}{b\left(b+2c\right)^2}\)+\(\dfrac{b}{c\left(c+2a\right)^2}\)+\(\dfrac{c}{a\left(a+2b\right)^2}\)≥\(\dfrac{4}{3}\)