Cho a,b,c dương. CMR
\(\dfrac{a^6}{b^3}+\dfrac{b^6}{c^3}+\dfrac{c^6}{a^3}\ge\dfrac{a^4}{c}+\dfrac{b^4}{a}+\dfrac{c^4}{b}\)
1. Cho a,b,c t/m: \(\left\{{}\begin{matrix}a\ge\dfrac{4}{3}\\b\ge\dfrac{4}{3}\\c\ge\dfrac{4}{3}\end{matrix}\right.\) và \(a+b+c=6\)
\(CMR:\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}+\dfrac{c}{c^2+1}\ge\dfrac{6}{5}\)
2. Cho x,y >0 t/m: \(2x+3y-13\ge0\)
Tìm min \(P=x^2+3x+\dfrac{4}{x}+y^2+\dfrac{9}{y}\)
Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)
\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)
CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)
Cộng vế theo vế:
\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)
Dấu \("="\Leftrightarrow a=b=c=2\)
Cho \(a,b,c>0\). CMR \(\sqrt{\dfrac{a}{b+c}}+\sqrt[3]{\dfrac{b}{c+a}}+\sqrt[4]{\dfrac{c}{a+b}}\ge\dfrac{7}{12}\cdot2^{\dfrac{6}{7}}\cdot3^{\dfrac{4}{7}}\)
Nhìn người hỏi là biết bài này khó rồi. Không liên quan nhưng anh Thắng đẹp zai làm giúp em bài này :)) https://hoc24.vn/hỏi-đáp/question/592811.html
Cho 3 số dương a,b,c.
CMR : \(\dfrac{a}{\sqrt{a^2+15bc}}+\dfrac{b}{\sqrt{b^2+15ac}}+\dfrac{c}{\sqrt{c^2+15ab}}\ge\dfrac{3}{4}\)
\(\dfrac{a}{\sqrt{a^2+15bc}}+\dfrac{b}{\sqrt{b^2+15ca}}+\dfrac{c}{\sqrt{c^2+15ab}}\ge\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{a^2}{a\sqrt{a^2+15bc}}+\dfrac{b^2}{b\sqrt{b^2+15ca}}+\dfrac{c^2}{c\sqrt{c^2+15ab}}\ge\dfrac{3}{4}\)
Áp dụng BĐT Caushy-Schwarz ta được:
\(\dfrac{a^2}{a\sqrt{a^2+15bc}}+\dfrac{b^2}{b\sqrt{b^2+15ca}}+\dfrac{c^2}{c\sqrt{c^2+15ab}}\ge\dfrac{\left(a+b+c\right)^2}{a\sqrt{a^2+15bc}+b\sqrt{b^2+15ca}+c\sqrt{c^2+15ab}}\)
Ta chứng minh rằng:
\(a\sqrt{a^2+15bc}+b\sqrt{b^2+15ca}+c\sqrt{c^2+15ab}\le\dfrac{4}{3}\left(a+b+c\right)^2\)
\(\Leftrightarrow\sqrt{a}\sqrt{a^3+15abc}+\sqrt{b}\sqrt{b^3+15abc}+\sqrt{c}\sqrt{c^3+15abc}\le\dfrac{4}{3}\left(a+b+c\right)^2\)
Áp dụng BĐT Bunhiacopxki ta được:
\(\sqrt{a}\sqrt{a^3+15abc}+\sqrt{b}\sqrt{b^3+15abc}+\sqrt{c}\sqrt{c^3+15abc}\le\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+45abc\right)}\)Ta tiếp tục chứng minh:
\(\dfrac{16}{9}\left(a+b+c\right)^3\ge a^3+b^3+c^3+45abc\)
\(\Leftrightarrow\dfrac{16}{9}\left(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\right)\ge a^3+b^3+c^3+45abc\)
Áp dụng BĐT AM-GM (Caushy) ta được:
\(\dfrac{16}{9}\left(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\right)\ge\dfrac{16}{9}\left(a^3+b^3+c^3+3.2\sqrt{ab}.2.\sqrt{bc}.2.\sqrt{ca}\right)=\dfrac{16}{9}.\left(a^3+b^3+c^3+24abc\right)\)
Ta chứng minh:
\(\dfrac{16}{9}\left(a^3+b^3+c^3+24abc\right)\ge a^3+b^3+c^3+45abc\)
\(\Leftrightarrow\dfrac{16}{9}a^3+\dfrac{16}{9}b^3+\dfrac{16}{9}c^3+\dfrac{16}{9}.24abc\ge a^3+b^3+c^3+45abc\)
\(\Leftrightarrow\dfrac{7}{9}\left(a^3+b^3+c^3\right)\ge\dfrac{7}{3}abc\) (*)
Áp dụng BĐT AM-GM (Caushy) ta được:
\(\dfrac{7}{9}\left(a^3+b^3+c^3\right)\ge\dfrac{7}{9}.3\sqrt[3]{a^3b^3c^3}=\dfrac{7}{3}abc\)
\(\Rightarrow\) (*) đúng.
Vậy BĐT đã được chứng minh. Dấu "=" xảy ra khi \(a=b=c>0\).
cho a,b,c là các số thực dương. Cmr
\(\dfrac{a^4}{b^3\left(c+a\right)}+\dfrac{b^4}{c^3\left(a+b\right)}+\dfrac{c^4}{a^3\left(b+c\right)}\ge\dfrac{3}{2}\)
Lời giải:
Áp dụng BĐT Cauchy ta có:
\(\frac{a^4}{b^3(c+a)}+\frac{c+a}{4a}+\frac{1}{2}\geq 3\sqrt[3]{\frac{a^3}{8b^3}}=\frac{3a}{2b}\)
\(\frac{b^4}{c^3(a+b)}+\frac{a+b}{4b}+\frac{1}{2}\geq 3\sqrt[3]{\frac{b^3}{8c^3}}=\frac{3b}{2c}\)
\(\frac{c^4}{a^3(b+c)}+\frac{b+c}{4c}+\frac{1}{2}\geq 3\sqrt[3]{\frac{c^3}{8a^3}}=\frac{3c}{2a}\)
Cộng theo vế và rút gọn:
\(\Rightarrow \frac{a^4}{b^3(c+a)}+\frac{b^4}{c^3(a+b)}+\frac{c^4}{a^3(b+c)}+\frac{1}{4}(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})+\frac{9}{4}\geq \frac{3}{2}(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\)
\(\Rightarrow \frac{a^4}{b^3(c+a)}+\frac{b^4}{c^3(a+b)}+\frac{c^4}{a^3(b+c)}\geq \frac{5}{4}(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})-\frac{9}{4}\geq \frac{5}{4}.3\sqrt[3]{\frac{a}{b}.\frac{b}{c}.\frac{c}{a}}-\frac{9}{4}\)
hay \( \frac{a^4}{b^3(c+a)}+\frac{b^4}{c^3(a+b)}+\frac{c^4}{a^3(b+c)}\geq \frac{5}{4}.3-\frac{9}{4}=\frac{3}{2}\)
Ta có đpcm
Dấu bằng xảy ra khi \(a=b=c\)
Cách khác:
Áp dụng BĐT Cauchy-Schwarz:
\(\text{VT}=\frac{(\frac{a^2}{b})^2}{b(c+a)}+\frac{(\frac{b^2}{c})^2}{c(a+b)}+\frac{(\frac{c^2}{a})^2}{a(b+c)}\geq \frac{\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)^2}{b(c+a)+c(a+b)+a(b+c)}\)
Tiếp tục áp dụng BĐT Cauchy-Schwarz:
\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq \frac{(a+b+c)^2}{b+c+a}=a+b+c\)
\(\Rightarrow \left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)^2\geq (a+b+c)^2\)
Do đó: \(\text{VT}\geq \frac{(a+b+c)^2}{2(ab+bc+ac)}\)
Theo hệ quả quen thuộc của BĐT Cauchy: \((a+b+c)^2\geq 3(ab+bc+ac)\)
Suy ra: \(\text{VT}\geq \frac{3(ab+bc+ac)}{2(ab+bc+ac)}=\frac{3}{2}\) (đpcm)
cho a,b,c dương thỏa a+b+c=3 chứng minh rằng
\(\dfrac{a}{b^3+16}+\dfrac{b}{c^3+16}+\dfrac{c}{a^3+16}\ge\dfrac{1}{6}\)
cho 3 số a,b,c dương và a+b+c=1.CMR
\(\dfrac{ab}{a^2+b^2}+\dfrac{bc}{b^2+c^2}+\dfrac{ca}{c^2+a^2}+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{15}{4}\)
Cho \(a;b;c\ge\dfrac{4}{3}\) thỏa mãn \(a+b+c=6\)
Tìm min: \(A=\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}+\dfrac{c}{c^2+1}\)
Lời giải:
Xét:
$\frac{a}{a^2+1}-\left(\frac{16}{25}-\frac{3}{25}a\right)=\frac{(a-2)^2(3a-4)}{25(a^2+1)}\geq 0$ với mọi $a\geq \frac{4}{3}$
$\Rightarrow \frac{a}{a^2+1}\geq \frac{16}{25}-\frac{3}{25}a$
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế, suy ra:
$A\geq \frac{48}{25}-\frac{3}{25}(a+b+c)=\frac{6}{5}$
Vậy $A_{\min}=\frac{6}{5}$.
Giá trị này đạt tại $a=b=c=2$
Cho \(a,b,c\) là các số dương . \(CMR\) \(\dfrac{a^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{b^3}{\left(b+c\right)\left(c+a\right)}+\dfrac{c^3}{\left(c+a\right)\left(a+b\right)}\ge\dfrac{1}{4}\left(a+b+c\right)\)
\(\dfrac{a^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge3\sqrt[3]{\dfrac{a^3\left(a+b\right)\left(b+c\right)}{64}}=\dfrac{3a}{4}\)
Tương tự:
\(\dfrac{b^3}{\left(b+c\right)\left(c+a\right)}+\dfrac{b+c}{8}+\dfrac{c+a}{8}\ge\dfrac{3b}{4}\)
\(\dfrac{c^3}{\left(c+a\right)\left(a+b\right)}+\dfrac{c+a}{8}+\dfrac{a+b}{8}\ge\dfrac{3c}{4}\)
Cộng vế:
\(VT+\dfrac{4\left(a+b+c\right)}{8}\ge\dfrac{3\left(a+b+c\right)}{4}\)
\(\Rightarrow VT\ge\dfrac{a+b+c}{4}\)
Dấu "=" xảy ra khi \(a=b=c\)
Cho a,b,c >0; a+b+c=6. Chứng minh rằng
\(\dfrac{a+b}{c^2+4}+\dfrac{b+c}{a^2+4}+\dfrac{c+a}{b^2+4}\ge\dfrac{3}{2}\)