Cho a,b,c là các số thực dương.CMR:
\(\dfrac{a^2}{a^2+b^2+c^2-bc}+\dfrac{b^2}{a^2+b^2+c^2-ca}+\dfrac{c^2}{a^2+b^2+c^2-ca}\le\dfrac{3}{2}\)
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AH
Akai Haruma
Giáo viên
12 tháng 11 2017
Lời giải:
Áp dụng BĐT Cauchy_ Schwarz ta có:
\(\text{VT}=\frac{a^6}{a^3+a^2b+ab^2}+\frac{b^6}{b^3+b^2c+bc^2}+\frac{c^6}{c^3+c^2a+ca^2}\)
\(\geq \frac{(a^3+b^3+c^3)^2}{a^3+a^2b+ab^2+b^3+b^2c+bc^2+c^3+c^2a+ca^2}\)
\(\Leftrightarrow \text{VT}\geq \frac{(a^3+b^3+c^3)^2}{a^3+b^3+c^3+ab(a+b)+bc(b+c)+ac(a+c)}\) (I)
Áp dụng BĐT Am-Gm ta có:
\(\left\{\begin{matrix} a^3+a^3+b^3\geq 3a^2b\\ b^3+b^3+c^3\geq 3b^2c\\ c^3+c^3+a^3\geq 3c^2a\end{matrix}\right.\Rightarrow 3(a^3+b^3+c^3)\geq 3(a^2b+b^2c+c^2a)\)
\(\Leftrightarrow a^3+b^3+c^3\geq a^2b+b^2c+c^2a\) (1)
Tương tự:
\(\left\{\begin{matrix} a^3+b^3+b^3\geq 3ab^2\\ b^3+c^3+c^3\geq 3bc^2\\ c^3+a^3+a^3\geq 3ca^2\end{matrix}\right.\Rightarrow 3(a^3+b^3+c^3)\geq 3(ab^2+bc^2+ca^2)\)
\(\Leftrightarrow a^3+b^3+c^3\geq ab^2+bc^2+ca^2(2)\)
Từ \((1);(2)\Rightarrow 2(a^3+b^3+c^3)\geq ab(a+b)+bc(b+c)+ac(c+a)\)
\(\Rightarrow a^3+b^3+c^3+ab(a+b)+bc(b+c)+ac(c+a)\leq 3(a^3+b^3+c^3)\) (II)
Từ \((I);(II)\Rightarrow \text{VT}\geq \frac{(a^3+b^3+c^3)^2}{a^3+b^3+c^3+ab(a+b)+bc(b+c)+ac(a+c)}\geq \frac{(a^3+b^3+c^3)^2}{3(a^3+b^3+c^3)}\)
\(\Leftrightarrow \text{VT}\geq \frac{a^3+b^3+c^3}{3}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c\)
16 tháng 10 2017
Từ \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow a+b+c\ge\dfrac{3\left(ab+bc+ca\right)}{a+b+c}\). Tức cần chứng minh
\(\dfrac{a^3}{b^2-bc+c^2}+\dfrac{b^3}{c^2-ac+a^2}+\dfrac{c^3}{a^2-ab+b^2}\ge a+b+c\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a^4}{ab^2-abc+ac^2}+\dfrac{b^4}{bc^2-abc+a^2b}+\dfrac{c^4}{a^2c-abc+b^2c}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^2b+a^2b+b^2c+bc^2+c^2a+ca^2-3abc}\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2\ge\left(a+b+c\right)\left(a^2b+a^2b+b^2c+bc^2+c^2a+ca^2-3abc\right)\)
\(\Leftrightarrow a^4+b^4+c^4+abc\left(a+b+c\right)\ge ab\left(a^2+b^2\right)+bc\left(b^2+c^2\right)+ca\left(c^2+a^2\right)\)
Đúng theo Schur bậc 4
26 tháng 10 2021
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 \("="\Leftrightarrow a=b=c=3\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 1 2021
\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)
\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)
\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 4 2021
\(\Leftrightarrow\dfrac{a}{\sqrt{4b^2+bc+4c^2}}+\dfrac{b}{\sqrt{4c^2+ca+4a^2}}+\dfrac{c}{\sqrt{4a^2+ab+4b^2}}\ge1\)
Ta có:
\(\sum\left(\dfrac{a}{\sqrt{4b^2+bc+4c^2}}\right)^2\sum a\left(4b^2+bc+4c^2\right)\ge\left(a+b+c\right)^3\)
Nên ta chỉ cần chứng minh:
\(\dfrac{\left(a+b+c\right)^3}{a\left(4b^2+bc+4c^2\right)+b\left(4c^2+ac+4a^2\right)+c\left(4a^2+ab+4b^2\right)}\ge1\)
\(\Leftrightarrow\dfrac{\left(a+b+c\right)^3}{4a\left(b^2+c^2\right)+4b\left(c^2+a^2\right)+4c\left(a^2+b^2\right)+3abc}\ge1\)
\(\Leftrightarrow a^3+b^3+c^3+3abc\ge ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\) (đúng theo Schur bậc 3)
23 tháng 9 2021
Áp dụng BĐT Cosi:
\(\dfrac{a}{\sqrt{b^2+ab}}=\dfrac{a\sqrt{2}}{\sqrt{2\left(b^2+ab\right)}}=\dfrac{a\sqrt{2}}{\sqrt{2b\left(a+b\right)}}\ge\dfrac{a\sqrt{2}}{\dfrac{2b+a+b}{2}}=\dfrac{2\sqrt{2}a}{a+3b}\)
Cmtt: \(\dfrac{b}{\sqrt{c^2+bc}}\ge\dfrac{2\sqrt{2}b}{b+3c};\dfrac{c}{\sqrt{a^2+ca}}\ge\dfrac{2\sqrt{2}c}{c+3a}\)
\(\Leftrightarrow P\ge2\sqrt{2}\left(\dfrac{a}{a+3b}+\dfrac{b}{b+3c}+\dfrac{c}{c+3a}\right)\\ \Leftrightarrow\dfrac{P}{\sqrt{2}}\ge2\left(\dfrac{a}{a+3b}+\dfrac{b}{b+3c}+\dfrac{c}{c+3a}\right)\\ \Leftrightarrow\dfrac{P}{\sqrt{2}}\ge\dfrac{2\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\dfrac{1}{3}\left(a+b+c\right)^2}\\ \Leftrightarrow\dfrac{P}{\sqrt{2}}\ge\dfrac{2}{\dfrac{4}{3}}=\dfrac{3}{2}\\ \Leftrightarrow P\ge\dfrac{3\sqrt{2}}{2}\)
Dấu \("="\Leftrightarrow a=b=c\)
Đặt vế trái của BĐT cần chứng minh là P
Ta có:
\(\dfrac{a^2}{a^2+b^2+c^2-bc}=\dfrac{2a^2}{2a^2+b^2+c^2+\left(b-c\right)^2}\le\dfrac{2a^2}{2a^2+b^2+c^2}=\dfrac{2a^2}{a^2+b^2+a^2+c^2}\)
\(\le\dfrac{1}{2}\left(\dfrac{a^2}{a^2+b^2}+\dfrac{a^2}{a^2+c^2}\right)\)
Tương tự:
\(\dfrac{b^2}{a^2+b^2+c^2-ac}\le\dfrac{1}{2}\left(\dfrac{b^2}{a^2+b^2}+\dfrac{b^2}{b^2+c^2}\right)\)
\(\dfrac{c^2}{a^2+b^2+c^2-ab}\le\dfrac{1}{2}\left(\dfrac{c^2}{a^2+c^2}+\dfrac{c^2}{b^2+c^2}\right)\)
Cộng vế với vế:
\(P\le\dfrac{1}{2}\left(\dfrac{a^2+b^2}{a^2+b^2}+\dfrac{b^2+c^2}{b^2+c^2}+\dfrac{c^2+a^2}{a^2+c^2}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)