cho a,b > 0 thỏa mãn \(a+b\ge4\) . Tìm GTNN của
\(\sqrt{9+a^2b^2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
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bài 1
ÁP dụng AM-GM ta có:
\(\frac{a^3}{b\left(2c+a\right)}+\frac{2c+a}{9}+\frac{b}{3}\ge3\sqrt[3]{\frac{a^3.\left(2c+a\right).b}{b\left(2c+a\right).27}}=a.\)
tương tự ta có:\(\frac{b^3}{c\left(2a+b\right)}+\frac{2a+b}{9}+\frac{c}{3}\ge b,\frac{c^3}{a\left(2b+c\right)}+\frac{2b+c}{9}+\frac{a}{3}\ge c\)
công tất cả lại ta có:
\(P+\frac{2a+b}{9}+\frac{2b+c}{9}+\frac{2c+a}{9}+\frac{a+b+c}{3}\ge a+b+c\)
\(P+\frac{2\left(a+b+c\right)}{3}\ge a+b+c\)
Thay \(a+b+c=3\)vào ta được":
\(P+2\ge3\Leftrightarrow P\ge1\)
Vậy Min là \(1\)
dấu \(=\)xảy ra khi \(a=b=c=1\)
p \(\ge\)\(\frac{4}{a^2+b^2+2\left(a+b\right)}\) +\(\sqrt{\left(1+ab\right)^2}\) (bunhia và cosi)
=\(\frac{4}{a^2+b^2+2ab}+1+ab=\frac{4}{\left(a+b\right)^2}+a+b+1\)
do \(a+b=ab\le\frac{\left(a+b\right)^2}{4}\Rightarrow a+b\ge4\)
dạt a+b = t thì t>=4
cần tìm min \(\frac{4}{t^2}+t+1=\frac{4}{t^2}+\frac{t}{16}+\frac{t}{16}+\frac{7t}{8}+1\)
\(\ge3.\sqrt[3]{\frac{4}{t^2}.\frac{t}{16}.\frac{t}{16}}+\frac{7.4}{8}+1=\frac{21}{4}\)
dau = xay ra khi a=b=2
Tìm GTNN a: $F= 14(a^2+b^2+c^2) + \dfrac{ab+bc+ca}{a^2b+b^2c+c^2a}$ | HOCMAI Forum - Cộng đồng học sinh Việt Nam
Ta có:
\(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}=\frac{1}{3}\)
\(\Leftrightarrow\left(a^2b+b^2c+c^2a\right)^2\le\left(a^2+b^2+c^2\right)\left(a^2b+b^2c+c^2a\right)\le\frac{\left(a^2+b^2+c^2\right)^3}{3}\le\left(a^2+b^2+c^2\right)^4\)
\(\Rightarrow a^2b+b^2c+c^2a\le\left(a^2+b^2+c^2\right)^2\)
Ta lại có:
\(ab+bc+ca=\frac{1-\left(a^2+b^2+c^2\right)^2}{2}\)
Làm tiếp.
Ta có
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}.\frac{3}{\sqrt[3]{abc}}\ge9\)
Dấu = xảy ra khi \(a=b=c=\frac{2014}{6}=\frac{1007}{3}\)
\(P^2=\left(9+a^2b^2\right)\left(\frac{1}{a}+\frac{1}{b}\right)^2=\left(\frac{3}{a}+\frac{3}{b}\right)^2+\left(a+b\right)^2\)
\(P^2\ge\left(\frac{12}{a+b}\right)^2+\left(a+b\right)^2=\frac{144}{\left(a+b\right)^2}+\frac{9\left(a+b\right)^2}{16}+\frac{7\left(a+b\right)^2}{16}\)
\(P^2\ge2\sqrt{\frac{144.9}{16}}+\frac{7.4^2}{16}=25\)
\(\Rightarrow P\ge5\)
Đặt P=\(\sqrt{9+a^2b^2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(=\sqrt{9\left(\frac{1}{a}+\frac{1}{b}\right)^2+a^2b^2\left(\frac{1}{a}+\frac{1}{b}\right)^2}\)
\(=\sqrt{\left(\frac{3}{a}+\frac{3}{b}\right)^2+\left(a+b\right)^2}\)
Theo cauchy-schwartz:
\(\left(\left(\frac{3}{a}+\frac{3}{b}\right)^2+\left(a+b\right)^2\right)\left(\left(\frac{3}{4}\right)^2+1^2\right)\ge\left[\frac{9}{4}\left(\frac{1}{a}+\frac{1}{b}\right)+a+b\right]^2\)
\(\frac{9}{4}\left(\frac{1}{a}+\frac{1}{b}\right)+a+b\ge\frac{9}{4}.\frac{4}{a+b}+a+b=\frac{9}{a+b}+a+b\)
Theo AM-GM:
\(\frac{9}{a+b}+a+b=a+b+\frac{16}{a+b}-\frac{7}{a+b}\ge2\sqrt{\left(a+b\right)\frac{16}{a+b}}-\frac{7}{a+b}\)
Mà a+b≥4
\(\Rightarrow\frac{9}{a+b}+a+b\ge2\sqrt{16}-\frac{7}{4}=\frac{25}{4}\)
=>P2≥\(\frac{\left(\frac{25}{4}\right)^2}{\left(\frac{3}{4}\right)^2+1^2}=5^2\)
=>P≥5
Dấu bằng xảy ra khi a=b=2
Vậy minP=5 khi a=b=2