Cho a,b,c dương và a^2+ b^2+c^2 = 12. Tìm GTLN của S = \(a\sqrt[3]{b^2+c^2}\) + \(b\sqrt[3]{c^2+a^2}\) + \(c\sqrt[3]{a^2+b^2}\)
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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 \(a=b=c=1\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 3 2023
Ta có:
\(ab+bc+ca\le\dfrac{1}{3}\left(a+b+c\right)^2=3\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+3}}\le\dfrac{a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)
Tương tự:
\(\dfrac{b}{\sqrt{b^2+3}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\) ; \(\dfrac{c}{\sqrt{c^2+3}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{b+c}\right)\)
Cộng vế:
\(P\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{c}{a+c}+\dfrac{a}{a+c}\right)=\dfrac{3}{2}\)
\(P_{max}=\dfrac{3}{2}\) khi \(a=b=c=1\)
6 tháng 5 2021
Ta có \(\sqrt{1+a^2}+\sqrt{2a}\le\sqrt{2\left(1+a^2+2a\right)}=\sqrt{2}\left(a+1\right)\).
Tương tự \(\sqrt{1+b^2}+\sqrt{2b}\le\sqrt{2}\left(b+1\right)\); \(\sqrt{1+c^2}+\sqrt{2c}\le\sqrt{2}\left(c+1\right)\).
Lại có \(\left(2-\sqrt{2}\right)\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\le\left(2-\sqrt{2}\right)\sqrt{3\left(a+b+c\right)}\le3\left(2-\sqrt{2}\right)\).
Do đó \(B\le\sqrt{2}\left(a+b+c+3\right)+3\left(2-\sqrt{2}\right)\le6\sqrt{2}+6-3\sqrt{2}=3\sqrt{2}+6\).
Dấu "=" xảy ra khi a = b = c = 1.
9 tháng 8 2020
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
N
25 tháng 12 2019
We have:
\(M=1-\frac{1}{3}\Sigma_{cyc}\frac{a^2+b^2}{a^2+b^2+3}\)
Consider:
\(\Sigma_{cyc}\frac{a^2+b^2}{a^2+b^2+3}\ge\frac{3}{2}\)
\(VT\ge\frac{\left(\Sigma_{cyc}\sqrt{a^2+b^2}\right)^2}{2\left(a^2+b^2+c^2\right)+9}\)
Prove:
\(\frac{\left(\Sigma_{cyc}\sqrt{a^2+b^2}\right)^2}{2\left(a^2+b^2+c^2\right)+9}\ge\frac{3}{2}\)
\(\Leftrightarrow4\Sigma_{cyc}\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}\ge2\left(a^2+b^2+c^2\right)+27\)
Consider:
\(\Sigma_{cyc}\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}\ge\Sigma_{cyc}a^2+\Sigma_{cyc}ab\)
\(\Rightarrow4\Sigma_{cyc}\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}\ge4\Sigma_{cyc}a^2+4\Sigma_{cyc}ab\)
Now we need to prove:
\(4\Sigma_{cyc}a^2+4\Sigma_{cyc}ab=2\Sigma_{cyc}a^2+27\)
\(\Leftrightarrow2\left(a+b+c\right)^2=27\) (not fail)
\(\Rightarrow M\le\frac{1}{2}\)
Sign '=' happen when \(a=b=c=\sqrt{\frac{3}{2}}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 4 2022
Ta có:
\(\left(a^2+b+c\right)\left(1+b+c\right)\ge\left(a+b+c\right)^2\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+b+c}}\le\dfrac{a\sqrt{1+b+c}}{a+b+c}\)
Tương tự: \(\dfrac{b}{\sqrt{b^2+a+c}}\le\dfrac{b\sqrt{1+c+a}}{a+b+c}\) ; \(\dfrac{c}{\sqrt{c^2+b+a}}\le\dfrac{c\sqrt{1+a+b}}{a+b+c}\)
Cộng vế:
\(P\le\dfrac{a\sqrt{1+b+c}+b\sqrt{1+c+a}+c\sqrt{1+a+b}}{a+b+c}\)
Lại có:
\(a\sqrt{1+b+c}+b\sqrt{1+c+a}+c\sqrt{1+a+b}\)
\(=\sqrt{a}.\sqrt{a+ab+ac}+\sqrt{b}.\sqrt{b+bc+ab}+\sqrt{c}.\sqrt{c+ac+bc}\)
\(\le\sqrt{\left(a+b+c\right)\left(a+b+c+2ab+2bc+2ca\right)}\)
\(\Rightarrow P\le\dfrac{\sqrt{\left(a+b+c\right)\left(a+b+c+2ab+bc+ca\right)}}{a+b+c}=\sqrt{\dfrac{a+b+c+2ab+2bc+2ca}{a+b+c}}\)
Do đó ta chỉ cần chứng minh:
\(\dfrac{a+b+c+2ab+2bc+2ca}{a+b+c}\le3\Leftrightarrow a+b+c\ge ab+bc+ca\)
Thật vậy:
\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)=\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)\ge\left(ab+bc+ca\right)^2\)
\(\Rightarrow a+b+c\ge ab+bc+ca\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
S
8 tháng 2 2020
\(B=\frac{a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\)
\(B=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(b+c\right)\left(c+a\right)}}\)
\(B\le\frac{\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{b+c}+\frac{c}{c+a}}{2}=\frac{3}{2}\Rightarrow B_{max}=\frac{3}{2}\)
\(\text{Dấu "=" xảy ra khi và chỉ khi:}a=b=c=\frac{1}{\sqrt{3}}\)
Ta có 4S=\(a\sqrt[3]{8.8.\left(b^2+c^2\right)}+b\sqrt[3]{8.8.\left(c^2+a^2\right)}+c.\sqrt[3]{8.8\left(a^2+b^2\right)}\)
Áp dụng BĐT cô-si, ta có \(\sqrt[3]{8.8.\left(b^2+c^2\right)}\le\frac{8+8+b^2+c^2}{3}\Rightarrow a\sqrt[3]{8.8.\left(b^2+c^2\right)}\le\frac{16}{3}a+\frac{1}{3}\left(ab^2+ac^2\right)\)
Tương tự, rồi cộng lại, ta có
\(4S\le\frac{16}{3}\left(a+b+c\right)+\frac{1}{3}\left(ab^2+ac^2+bc^2+ba^2+ca^2+cb^2\right)\)
Mà \(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=6\)
\(ab^2+bc^2+ca^2+a^2b+b^2c+c^2a\le\frac{2}{3}\left(a+b+c\right)\left(a^2+b^2+c^2\right)\le\frac{2}{3}.6.12=48\)
=> \(4S\le\frac{16}{3}.6+\frac{1}{3}.48=48\Rightarrow S\le12\)
Dấu = xảy ra <=> a=b=c=2
^_^
bài này dùng holder 3 số là nhanh nhất