Cho a,b,c > 0 có a+b+c = 3 Tìm gtln của
\(Q=\dfrac{ab}{\sqrt{3a^2+b^2}+1}+\dfrac{bc}{\sqrt{3b^2+c^2}+1}+\dfrac{ca}{\sqrt{3c^2+a^2}+1}\)
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NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 3 2023
\(Q=\dfrac{2a}{\sqrt{a^2+ab+bc+ca}}+\dfrac{b}{\sqrt{b^2+ab+bc+ca}}+\dfrac{c}{\sqrt{c^2+ab+bc+ca}}\)
\(=\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(=\sqrt{\dfrac{2a}{a+b}.\dfrac{2a}{a+c}}+\sqrt{\dfrac{2b}{a+b}.\dfrac{b}{2\left(b+c\right)}}+\sqrt{\dfrac{2c}{a+c}.\dfrac{c}{2\left(b+c\right)}}\)
\(\le\dfrac{1}{2}\left(\dfrac{2a}{a+b}+\dfrac{2a}{a+c}+\dfrac{2b}{a+b}+\dfrac{b}{2\left(b+c\right)}+\dfrac{2c}{a+c}+\dfrac{c}{2\left(b+c\right)}\right)\)
\(=\dfrac{9}{4}\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{7}{\sqrt{15}};\dfrac{1}{\sqrt{15}};\dfrac{1}{\sqrt{15}}\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 1 2021
\(a^2-ab+b^2=\dfrac{1}{4}\left(a+b\right)^2+\dfrac{3}{4}\left(a-b\right)^2\ge\dfrac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow P\le\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
AH
Akai Haruma
Giáo viên
27 tháng 8 2021
Lời giải:
Theo hệ quả quen thuộc của bđt AM-GM:
$(a+b+c)^2\leq 3(a^2+b^2+c^2)\leq 9$
$\Rightarrow a+b+c\leq 3$ (đpcm)
Từ đây ta có:
\(E\leq \frac{a}{\sqrt[3]{(a+b+c)a+bc}}+\frac{b}{\sqrt[3]{(a+b+c)b+ac}}+\frac{c}{\sqrt[3]{c(a+b+c)+ab}}\)
\(=\frac{a}{\sqrt[3]{(a+b)(a+c)}}+\frac{b}{\sqrt[3]{(b+c)(b+a)}}+\frac{c}{\sqrt[3]{(c+a)(c+b)}}\)
\(\leq \frac{\sqrt[3]{2}}{3}(\frac{a}{2}+\frac{a}{a+b}+\frac{a}{a+c})+\frac{\sqrt[3]{2}}{3}(\frac{b}{2}+\frac{b}{b+a}+\frac{b}{b+c})+\frac{\sqrt[3]{2}}{3}(\frac{c}{2}+\frac{c}{c+a}+\frac{c}{c+b})\)
\(=\frac{\sqrt[3]{2}(a+b+c)}{6}+\frac{\sqrt[3]{2}}{3}(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a})\leq \frac{3\sqrt[3]{2}}{2}\)
Vậy.................
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 8 2021
\(3\ge a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\Rightarrow a+b+c\le3\)
\(\Rightarrow\dfrac{a}{\sqrt[3]{3a+bc}}\le\dfrac{a}{\sqrt[3]{a\left(a+b+c\right)+bc}}=\sqrt[3]{2}.\sqrt[3]{\dfrac{a}{a+b}.\dfrac{a}{a+c}.\dfrac{a}{2}}\le\dfrac{\sqrt[3]{2}}{3}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{a}{2}\right)\)
Cộng vế và rút gọn:
\(E\le\dfrac{\sqrt[3]{2}}{3}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}+\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{a+b+c}{2}\right)\)
\(E\le\dfrac{\sqrt[3]{2}}{3}\left(3+\dfrac{3}{2}\right)=\dfrac{3\sqrt[3]{2}}{2}\)
28 tháng 5 2021
Áp dụng bđt Cauchy - Schwarz ta có:\(Q=\dfrac{2-2a^2b^2}{\left(1+a^2\right)\left(1+b^2\right)}+\dfrac{2}{\sqrt{1+c^2}}=\dfrac{2\left(1-ab\right)\left(1+ab\right)}{\left(ab+bc+ca+a^2\right)\left(ab+bc+ca+b^2\right)}+\dfrac{2}{\sqrt{1+c^2}}=\dfrac{2\left(bc+ca\right)\left(1+ab\right)}{\left(a+b\right)^2\left(b+c\right)\left(c+a\right)}+\dfrac{2}{\sqrt{1+c^2}}=\dfrac{2c\left(1+ab\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\dfrac{2c\left(1+ab\right)}{\sqrt{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}}+\dfrac{2}{\sqrt{1+c^2}}\le\dfrac{2c\left(1+ab\right)}{\sqrt{\left(ab+1\right)^2\left(c^2+1\right)}}+\dfrac{2}{\sqrt{1+c^2}}=\dfrac{2c}{\sqrt{c^2+1}}+\dfrac{2}{\sqrt{c^2+1}}=\dfrac{2\left(c+1\right)}{\sqrt{c^2+1}}\le\dfrac{2\left(c+1\right)}{\sqrt{\dfrac{\left(c+1\right)^2}{2}}}=2\sqrt{2}\)Dấu "=" xảy ra khi a = b = \(\sqrt{2}-1;c=1\).
Vậy..
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\((3a^2+b^2)(3+1)\geq (3a+b)^2\Rightarrow \sqrt{3a^2+b^2}\ge \frac{3a+b}{2}\)
\(\Rightarrow \frac{ab}{\sqrt{3a^2+b^2}+1}\leq \frac{2ab}{3a+b+2}\)
Thực hiện tương tự với các phân thức còn lại và cộng theo vế:
\(\Rightarrow Q\leq \frac{2ab}{3a+b+2}+\frac{2bc}{3b+c+2}+\frac{2ac}{3c+a+2}\)
\(\Leftrightarrow 3Q\leq \frac{6ab}{3a+b+2}+\frac{6bc}{3b+c+2}+\frac{6ac}{3c+a+2}\)
\(\Leftrightarrow 3Q\le 2b-\frac{2b^2+4b}{3a+b+2}+2c-\frac{2c^2+4c}{3b+c+2}+2a-\frac{2a^2+4a}{3c+a+2}\)
\(\Leftrightarrow 3Q\leq 6-\left(\frac{2b^2+4b}{3a+b+2}+\frac{2c^2+4c}{3b+c+2}+\frac{2a^2+4a}{3c+a+2}\right)(1)\)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{2b^2}{3a+b+2}+\frac{2c^2}{3b+c+2}+\frac{2a^2}{3c+a+2}\geq \frac{2(b+c+a)^2}{3a+b+2+3b+c+2+3c+a+2}=\frac{2(a+b+c)^2}{4(a+b+c)+6}=1(2)\)
Và:
\(\frac{4b}{3a+b+2}+\frac{4c}{3b+c+2}+\frac{4a}{3c+a+2}=4\left(\frac{b^2}{3ab+b^2+2b}+\frac{c^2}{3bc+c^2+2c}+\frac{a^2}{3ac+a^2+2a}\right)\)
\(\geq \frac{4(b+c+a)^2}{3ab+b^2+2b+3bc+c^2+3ac+a^2+2a}=\frac{4(a+b+c)^2}{(a+b+c)^2+2(a+b+c)+(ab+bc+ac)}\)
\(\geq \frac{4(a+b+c)^2}{(a+b+c)^2+2(a+b+c)+\frac{(a+b+c)^2}{3}}=2(3)\) (AM-GM)
Từ \((1); (2); (3)\Rightarrow 3Q\leq 6-(2+1)\Leftrightarrow 3Q\leq 3\Leftrightarrow Q\leq 1\)
Vậy Q(max) là $1$
Dấu bằng xảy ra khi \(a=b=c=1\)
Akai Haruma cô ơi làm giùm em với