Cho \(a,b,c>0\)
CMR: \(\frac{1+\sqrt{3}}{3\sqrt{3}}\left(a^2+b^2+c^2\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge a+b+c+\sqrt{a^2+b^2+c^2}\)
HELP ME
Cho \(a,b,c>0\)
CMR: \(\frac{1+\sqrt{3}}{3\sqrt{3}}\left(a^2+b^2+c^2\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge a+b+c+\sqrt{a^2+b^2+c^2}\)
Ủa @@ Mình vừa đăng câu trả lời rồi mà sao giờ không thấy nhỉ @@
Ta có: \(\left(a^2+b^2+c^2\right)\ge\frac{1}{3}\left(a+b+c\right)^2\Rightarrow\sqrt{3}\sqrt{a^2+b^2+c^2}\ge a+b+c\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\ge\frac{9}{\sqrt{3}\sqrt{a^2+b^2+c^2}}=\frac{3\sqrt{3}}{\sqrt{a^2+b^2+c^2}}\)
\(\Rightarrow\frac{1}{3}\left(a^2+b^2+c^2\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\frac{1}{3}.\frac{1}{3}\left(a+b+c\right)^2.\frac{9}{a+b+c}=a+b+c\) \(\left(1\right)\)
và \(\left(a^2+b^2+c^2\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(a^2+b^2+c^2\right).\frac{3\sqrt{3}}{\sqrt{a^2+b^2+c^2}}\)
\(\Rightarrow\frac{1}{3\sqrt{3}}\left(a^2+b^2+c^2\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\sqrt{a^2+b^2+c^2}\) \(\left(2\right)\)
Cộng vế với vế của (1) và (2) ta có đpcm
Dấu "=" xảy ra khi a=b=c
\(\left(a^2+b^2+c^2\right)\ge\frac{1}{3}\left(a+b+c\right)\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0.\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}=\frac{3}{\sqrt[3]{abc}}\ge\frac{3}{\frac{a+b+c}{3}}=\frac{9}{a+b+c}\)
Bài 1 :Cho a,b,c dương thỏa mãn a+b+c=2
CMR \(\frac{bc}{\sqrt{3a^2+4}}+\frac{ca}{\sqrt{3b^2+4}}+\frac{ab}{\sqrt{3c^2+4}}\ge\frac{\sqrt{3}}{3}\)
Bài 2:Cho a,b,c>0. CMR
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)
\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)
\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)
\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)
Từ ( 1 ) và ( 2 ) có đpcm
CMR: \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\) với mọi a,b,c >0
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}=\frac{c^2}{c^2\left(a+b\right)}+\frac{a^2}{a^2\left(b+c\right)}+\frac{b^2}{b^2\left(c+a\right)}+\frac{\left(\sqrt[3]{abc}\right)^2}{2abc}\)
Áp dụng BĐT Bun :
\(\frac{c^2}{c^2\left(a+b\right)}+\frac{a^2}{a^2\left(b+c\right)}+\frac{b^2}{b^2\left(a+c\right)}+\frac{\left(\sqrt[3]{abc}\right)^2}{2abc}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{c^2\left(a+b\right)+a^2\left(b+c\right)+b^2\left(a+c\right)+2abc}=...\)
Dấu ''='' xảy ra khi a = b =c
các bạn làm được ý nào thì làm ý đó nha
1. Cho a,b,c là độ dài 3 cạnh tam giác. Chứng minh:
a) \(\frac{1}{\left(a+b-c\right)^2}+\frac{1}{\left(a-b+c\right)^2}+\frac{1}{\left(b+c-a\right)^2}\ge\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
b) \(\frac{1}{\left(a+b-c\right)^3}+\frac{1}{\left(a-b+c\right)^3}+\frac{1}{\left(b+c-a\right)^3}\ge\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\)
c) \(\frac{1}{\left(a+b-c\right)^{200}}+\frac{1}{\left(a-b+c\right)^{200}}+\frac{1}{\left(b+c-a\right)^{200}}\ge\frac{1}{a^{200}}+\frac{1}{b^{200}}+\frac{1}{c^{200}}\)
d) \(\frac{1}{8}\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\sqrt{abc\left(-a+b+c\right)\left(a-b+c\right)\left(a+b-c\right)}\)
e) \(a+b+c< \sqrt{a\left(b+c\right)}+\sqrt{b\left(a+c\right)}+\sqrt{c\left(a+b\right)}\)
f) \(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}< \sqrt{6}\)
g) \(\sqrt{-a+b+c}+\sqrt{a-b+c}+\sqrt{a+b-c}\le\sqrt{3\left(a+b+c\right)}\)
Bài 1: Cho a,b,c là đọ dài 3 cạnh của một tam giác. CMR: \(\frac{1}{\sqrt{b+c-a}}+\frac{1}{\sqrt{a+c-b}}+\frac{1}{\sqrt{a+b-c}}\ge\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}.\)
Bài 2: Cho a,b,c >0. CMR: \(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(a+c-b\right).\)
Đặt ⎧⎪⎨⎪⎩a+b−c=xb+c−a=yc+a−b=z(x,y,z>0){a+b−c=xb+c−a=yc+a−b=z(x,y,z>0)
⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩a=z+x2b=x+y2c=y+z2⇒{a=z+x2b=x+y2c=y+z2
⇒√a(1b+c−a−1√bc)=√2(z+x)2(1y−2√(x+y)(y+z))≥√x+√z2(1y−2√xy+√yz)=√x+√z2y−1√y⇒a(1b+c−a−1bc)=2(z+x)2(1y−2(x+y)(y+z))≥x+z2(1y−2xy+yz)=x+z2y−1y
Tương tự
⇒∑√a(1b+c−a−1√bc)≥∑√x+√z2y−∑1√y⇒∑a(1b+c−a−1bc)≥∑x+z2y−∑1y
⇒VT≥∑[x√x(y+z)]2xyz−∑√xy√xyz≥2√xyz(x+y+z)2xyz−x+y+z√xyz≐x+y+z√xyz−x+y+z√xyz=0⇒VT≥∑[xx(y+z)]2xyz−∑xyxyz≥2xyz(x+y+z)2xyz−x+y+zxyz≐x+y+zxyz−x+y+zxyz=0
(∑√xy≤x+y+z,x√x(y+z)≥2x√xyz)(∑xy≤x+y+z,xx(y+z)≥2xxyz)
dấu = ⇔x=y=z⇔a=b=c
Mai Anh ! cậu giỏi quá, cậu nè :33
Ha~ Idol về mảng copy nay giỏi quá lè:33. Tác hại của việc copy paste là đây
Lần sai copy paste nhớ nhìn lại với chỉnh sửa đi nhá. Ko để này lộ liễu bôi bác lắm
Copy always mà vẫn 50k giải tuần đấy, ghê=))
1) Cho a,b,c>0 tm a+b+c=3. Cmr \(\frac{1}{2+a^2+b^2}+\frac{1}{2+b^2+c^2}+\frac{1}{2+c^2+a^2}\le\frac{3}{4}\)
2) Cho a,b,c>0 tm a^2+b^2+c^2 bé hơn hoặc bằng abc. Cmr \(\frac{a}{a^2+bc}+\frac{b}{b^2+ca}+\frac{c}{c^2+ab}\le\frac{1}{2}\)
3) Cho a,b,c>0 tm a+b+c<=3. Cmr \(\frac{ab}{\sqrt{3+c}}+\frac{bc}{\sqrt{3+a}}+\frac{ca}{\sqrt{3+b}}\le\frac{3}{2}\)
4) Cho a,b,c>0 tm a+b+c=2. Cmr \(\frac{a}{\sqrt{4a+3bc}}+\frac{b}{\sqrt{4b+3ca}}+\frac{c}{\sqrt{4c+3ab}}\le1\)
5) Cho a,b,c>0. Cmr \(\sqrt{\frac{a^3}{5a^2+\left(b+c\right)^2}}+\sqrt{\frac{b^3}{5b^2+\left(c+a\right)^2}}+\sqrt{\frac{c^3}{5c^2+\left(a+b\right)^2}}\le\sqrt{\frac{a+b+c}{3}}\)
6) Cho a,b,c>0. Cmr \(\frac{a^2}{\left(2a+b\right)\left(2a+c\right)}+\frac{b^2}{\left(2b+a\right)\left(2b+c\right)}+\frac{c^2}{\left(2c+a\right)\left(2c+b\right)}\le\frac{1}{3}\)
Giúp mình với nhé các bạn
Cho a,b,c>0 thỏa mãn: a.b.c=8
Chứng minh: \(\frac{a^2}{\sqrt{\left(1+a^3\right).\left(1+b^3\right)}}+\frac{b^2}{\sqrt{\left(1+b^3\right).\left(1+c^3\right)}}+\frac{c^2}{\sqrt{\left(1+c^3\right).\left(1+a^3\right)}}\ge\frac{4}{3}\)
Lời giải:
Áp dụng BĐT AM-GM:
\(a^3+1=(a+1)(a^2-a+1)\leq \left(\frac{a+1+a^2-a+1}{2}\right)^2=\left(\frac{a^2+2}{2}\right)^2\)
\(b^3+1\leq \left(\frac{b^2+2}{2}\right)^2\)
\(\Rightarrow \sqrt{(a^3+1)(b^3+1)}\leq \frac{(a^2+2)(b^2+2)}{4}\)
\(\Rightarrow \frac{a^2}{\sqrt{(a^3+1)(b^3+1)}}\geq \frac{4a^2}{(a^2+2)(b^2+2)}\)
Hoàn toàn tương tự với các phân thức còn lại:
\(\Rightarrow \text{VT}\geq \underbrace{\frac{4a^2}{(a^2+2)(b^2+2)}+\frac{4b^2}{(b^2+2)(c^2+2)}+\frac{4c^2}{(c^2+2)(a^2+2)}}_{M}\)
Ta cần CM \(M\geq \frac{4}{3}\)
\(\Leftrightarrow \frac{a^2(c^2+2)+b^2(a^2+2)+c^2(b^2+2)}{(a^2+2)(b^2+2)(c^2+2)}\geq \frac{1}{3}\)
\(\Leftrightarrow 3(a^2b^2+b^2c^2+c^2a^2)+6(a^2+b^2+c^2)\geq (a^2+2)(b^2+2)(c^2+2)\)
\(\Leftrightarrow 3(a^2b^2+b^2c^2+c^2a^2)+6(a^2+b^2+c^2)\geq (abc)^2+2(a^2b^2+b^2c^2+c^2a^2)+4(a^2+b^2+c^2)+8\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2(a^2+b^2+c^2)\geq 72\)
Điều này luôn đúng do theo BĐT AM-GM thì: \(\left\{\begin{matrix} a^2b^2+b^2c^2+c^2a^2\geq 3\sqrt[3]{(abc)^4}=3\sqrt[3]{8^4}=48\\ 2(a^2+b^2+c^2)\geq 6\sqrt[3]{(abc)^2}=6\sqrt[3]{8^2}=24\end{matrix}\right.\)
Do đó ta có đpcm
Dấu "=" xảy ra khi $a=b=c=2$
cho ba số thực dương a,b,c. cmr : \(\sqrt[3]{5a^2b+3}+\sqrt[3]{5b^2c+3}+\sqrt[3]{5c^2a+3}\le\frac{21}{12}\left(a+b+c\right)+\frac{1}{4}\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
help me!
cho a,b,c>0
Cm: \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}=\frac{c^2}{c^2(a+b)}+\frac{a^2}{a^2(b+c)}+\frac{b^2}{b^2(c+a)}+\frac{(\sqrt[3]{abc})^2}{2abc}\)
\(\geq \frac{(c+a+b+\sqrt[3]{abc})^2}{c^2(a+b)+a^2(b+c)+b^2(c+a)+2abc}=\frac{(a+b+c+\sqrt[3]{abc})^2}{(a+b)(b+c)(c+a)}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c$