cho a,b,c là các số dương thỏa mãn: a+b+c=1
cmr :\(\dfrac{ab+c}{c+1}+\dfrac{bc+a}{a+1}+\dfrac{ac+b}{b+1}\le1\)
Cho các số thực dương a,b,c thỏa mãn \(ab+bc+ca\ge3\) . CMR: \(\dfrac{1}{a^2+b^2+1}+\dfrac{1}{b^2+c^2+1}+\dfrac{1}{c^2+a^2+1}\le1\)
Lời giải:
Áp dụng BĐT Bunhiacopxky:
$(a^2+b^2+1)(1+1+c^2)\geq (a+b+c)^2$
$\Rightarrow \frac{1}{a^2+b^2+1}\leq \frac{c^2+2}{(a+b+c)^2}$
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế:
$\text{VT}\leq \frac{a^2+b^2+c^2+6}{(a+b+c)^2}=\frac{a^2+b^2+c^2+6}{a^2+b^2+c^2+2(ab+bc+ac)}\leq \frac{a^2+b^2+c^2+6}{a^2+b^2+c^2+2.3}=1$
Ta có đpcm.
Dấu "=" xảy ra khi $a=b=c=1$
cho ba số dương \(0\le a\le b\le c\le1\) CMR \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le2\)
Vì \(0\le a\le b\le c\le1\) nên:
\(\left(a-1\right)\left(b-1\right)\ge ab+1\ge a+b\Leftrightarrow\dfrac{1}{ab+1}\le\dfrac{1}{a+b}\Leftrightarrow\dfrac{c}{ab+1}\le\dfrac{c}{a+b}\left(1\right)\)
Tương tự: \(\dfrac{a}{bc+1}\le\dfrac{a}{b=c}\left(2\right);\dfrac{b}{ac+1}\le\dfrac{b}{a+c}\left(3\right)\)
Do đó: \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\left(4\right)\)
Mà: \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\le\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\left(5\right)\)
Từ (4) và (5) suy ra \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\left(đpcm\right)\)
a,b,c là các số thực dương thỏa mãn a+b+c=1. CMR: \(\dfrac{a+bc}{b+c}+\dfrac{b+ca}{c+a}+\dfrac{c+ab}{a+b}>=2\)
Lời giải:
$\text{VT}=\frac{a(a+b+c)+bc}{b+c}+\frac{b(a+b+c)+ac}{a+c}+\frac{c(a+b+c)+ab}{a+b}$
$=\frac{(a+b)(a+c)}{b+c}+\frac{(b+a)(b+c)}{a+c}+\frac{(c+a)(c+b)}{a+b}$
Áp dụng BĐT AM-GM:
$\frac{(a+b)(a+c)}{b+c}+\frac{(b+a)(b+c)}{a+c}\geq 2\sqrt{(a+b)^2}=2(a+b)$
$\frac{(b+c)(b+a)}{a+c}+\frac{(c+a)(c+b)}{a+b}\geq 2\sqrt{(b+c)^2}=2(b+c)$
$\frac{(a+b)(a+c)}{b+c}+\frac{(c+a)(c+b)}{a+b}\geq 2\sqrt{(c+a)^2}=2(a+c)$
Cộng các BĐT trên theo vế và thu gọn:
$\text{VT}\geq 2(a+b+c)=2$
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
Cho a, b, c là các số thực dương thỏa mãn \(\sqrt{a}+\sqrt{b}+\sqrt{c}=1\) . Cmr
\(\sqrt{\dfrac{ab}{a+b+2c}}+\sqrt{\dfrac{bc}{c+b+2a}}+\sqrt{\dfrac{ca}{a+c+2b}}\le\dfrac{1}{2}\)
Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=1\)
BĐT trở thành: \(\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}+\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}+\dfrac{zx}{\sqrt{x^2+z^2+2y^2}}\le\dfrac{1}{2}\)
Ta có:
\(x^2+z^2+y^2+z^2\ge\dfrac{1}{2}\left(x+z\right)^2+\dfrac{1}{2}\left(y+z\right)^2\ge\left(x+z\right)\left(y+z\right)\)
\(\Rightarrow\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}\le\dfrac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}\le\dfrac{1}{2}\left(\dfrac{xy}{x+z}+\dfrac{xy}{y+z}\right)\)
Tương tự: \(\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}\le\dfrac{1}{2}\left(\dfrac{yz}{x+y}+\dfrac{yz}{x+z}\right)\)
\(\dfrac{zx}{\sqrt{z^2+x^2+2y^2}}\le\dfrac{1}{2}\left(\dfrac{zx}{x+y}+\dfrac{zx}{y+z}\right)\)
Cộng vế với vế:
\(VT\le\dfrac{1}{2}\left(\dfrac{zx+yz}{x+y}+\dfrac{xy+zx}{y+z}+\dfrac{yz+xy}{z+x}\right)=\dfrac{1}{2}\left(x+y+z\right)=\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c\)
Cho a, b, c là các số dương thỏa mãn: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\). CMR: \(\dfrac{a^2}{a+bc}+\dfrac{b^2}{b+ca}+\dfrac{c^2}{c+ba}\le\dfrac{a+b+c}{4}\)
Sửa \(\le\) thành \(\ge\) nha bạn
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ca=abc\)
Ta có \(\dfrac{a^2}{a+bc}=\dfrac{a^3}{a^2+abc}=\dfrac{a^3}{a^2+ab+bc+ca}=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}\)
Tương tự: \(\left\{{}\begin{matrix}\dfrac{b^2}{b+ca}=\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}\\\dfrac{c^2}{c+ba}=\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}\end{matrix}\right.\)
Áp dụng BĐT cosi:
\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3}{4}a\)
\(\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge3\sqrt[3]{\dfrac{b^3}{64}}=\dfrac{3}{4}b\)
\(\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}+\dfrac{b+c}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{c^3}{64}}=\dfrac{3}{4}c\)
Cộng VTV:
\(\Leftrightarrow VT+\dfrac{a+b}{8}+\dfrac{a+c}{8}+\dfrac{b+c}{8}\ge\dfrac{3}{4}\left(a+b+c\right)\\ \Leftrightarrow VT\ge\dfrac{3\left(a+b+c\right)}{4}-\dfrac{2\left(a+b+c\right)}{8}\\ \Leftrightarrow VT\ge\dfrac{a+b+c}{4}\)
Dấu \("="\Leftrightarrow a=b=c=3\)
Cho các số thực dương \(a,b,c\) thỏa mãn : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\). Tìm giá trị lớn nhất của biểu thức :
\(P=\sqrt{\dfrac{a}{a+bc}}+\sqrt{\dfrac{b}{b+ac}}+\sqrt{\dfrac{c}{c+ab}}\)
Thỏa mãn $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ hay $a+b+c=1$ vậy bạn?
Cho a,b,c là các số thực dương thỏa mãn a+b+c=1. Chứng minhh rằng:
\(\left(1+ab+bc+ca\right)\left(\dfrac{1}{a+bc}+\dfrac{1}{b+ca}+\dfrac{1}{c+ab}\right)\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Leftrightarrow\left(1+ab+bc+ca\right)\left(\dfrac{1}{\left(a+b\right)\left(a+c\right)}+\dfrac{1}{\left(a+b\right)\left(b+c\right)}+\dfrac{1}{\left(a+c\right)\left(b+c\right)}\right)\le\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow\dfrac{2\left(1+ab+bc+ca\right)\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow\dfrac{2\left(1+ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{ab+bc+ca}{abc}\)
Áp dụng BĐT quen thuộc:
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\dfrac{8}{9}\left(ab+bc+ca\right)\left(a+b+c\right)=\dfrac{8}{9}\left(ab+bc+ca\right)\)
\(\Rightarrow\dfrac{2\left(1+ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9\left(1+ab+bc+ca\right)}{4\left(ab+bc+ca\right)}\)
Ta chỉ cần chứng minh:
\(\dfrac{9\left(1+ab+bc+ca\right)}{4\left(ab+bc+ca\right)}\le\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow4\left(ab+bc+ca\right)^2\ge9abc+9abc\left(ab+bc+ca\right)\)
Do \(3\left(ab+bc+ca\right)^2\ge9abc\left(a+b+c\right)=9abc\)
Nên ta chỉ cần chứng minh:
\(\left(ab+bc+ca\right)^2\ge9abc\left(ab+bc+ca\right)\)
\(\Leftrightarrow ab+bc+ca\ge9abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)
Hiển nhiên đúng do \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}=9\)
Cho a,b,c là các số thực dương thỏa mãn \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\). Tìm giá trị lớn nhất nhất của biểu thức: \(P=\dfrac{1}{\sqrt{a^2-ab+b^2}}+\dfrac{1}{\sqrt{b^2-bc+c^2}}+\dfrac{1}{\sqrt{c^2}-ac+a^2}\)
cái cuối là \(\dfrac{1}{\sqrt{c^2-ca+a^2}}\) nha
\(a^2+b^2-ab\ge\dfrac{1}{2}\left(a+b\right)^2-\dfrac{1}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\dfrac{1}{\sqrt{a^2-ab+b^2}}\le\dfrac{1}{\sqrt{\dfrac{1}{4}\left(a+b\right)^2}}=\dfrac{2}{a+b}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
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 \(a=b=c=1\)
Cho a, b, c là số thực dương thỏa mãn: a+b+c=1. Tìm GTLN của biểu thức: \(P=\sqrt{\dfrac{ab}{c+ab}}+\sqrt{\dfrac{bc}{a+bc}}+\sqrt{\dfrac{ac}{b+ac}}\)
Lời giải:
Áp dụng BĐT AM-GM:
\(P=\sum \sqrt{\frac{ab}{c+ab}}=\sum \sqrt{\frac{ab}{c(a+b+c)+ab}}=\sum \sqrt{\frac{ab}{(c+a)(c+b)}}\)
\(\leq \sum \frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)=\frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\)
Vậy $P_{\max}=\frac{3}{2}$ khi $a=b=c=\frac{1}{3}$