Question

[Phùng Minh Quân]

\(a,b,c,d\inℝ\) thoả mãn \(\left|a+b\right|\ge\left|c+d\right|\). CM : 

\(\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge2\left(\sqrt[4]{\left|a+b\right|^3\left|c+d\right|}-\sqrt[4]{\left|a+b\right|\left|c+d\right|^3}\right)\)

Answer 1

[Con Chim 7 Màu]

\(\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge\sqrt{\left(a+b\right)^2+\left(c+d\right)^2}\)

Cần CM : \(\sqrt{\left(a+b\right)^2+\left(c+d\right)^2}\ge\left|a+b\right|-\left|c+d\right|\)

\(\Leftrightarrow\)\(\left(a+b\right)^2+\left(c+d\right)^2\ge\left(a+b\right)^2+\left(c+d\right)^2-2\left|\left(a+b\right)\left(c+d\right)\right|\)

\(\Leftrightarrow\)\(\left|\left(a+b\right)\left(c+d\right)\right|\ge0\) ( luôn đúng \(\forall\left|a+b\right|\ge\left|c+d\right|\) ) 

Do đó \(VT\ge\left|a+b\right|-\left|c+d\right|=\left(\sqrt{\left|a+b\right|}\right)^2-\left(\sqrt{\left|c+d\right|}\right)^2\)

\(=\left(\sqrt{\left|a+b\right|}+\sqrt{\left|c+d\right|}\right)\left(\sqrt{\left|a+b\right|}-\sqrt{\left|c+d\right|}\right)\)

\(\ge2\sqrt[4]{\left|a+b\right|.\left|c+d\right|}\left(\sqrt{\left|a+b\right|}-\sqrt{\left|c+d\right|}\right)\)

\(=2\left(\sqrt[4]{\left|a+b\right|^3.\left|c+d\right|}-\sqrt[4]{\left|a+b\right|.\left|c+d\right|^3}\right)\) ( đpcm ) 

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Answer 2

[Trần Phúc Khang]

Áp dụng bất đẳng thức Mincoxki ta có 

\(\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge\sqrt{\left(a+b\right)^2+\left(c+d\right)^2}\)

Buniacoxki \(\sqrt{\left(\left(a+b\right)^2+\left(c+d\right)^2\right)\left(1+1\right)}\ge|a+b|+|c+d|\)

Khi đó cần Cm

\(|a+b|+|c+d|\ge2\left(\sqrt{|a+b|^3|c+d|}-\sqrt{|c+d|^3|a+b|}\right)\)

Đặt \(\sqrt[4]{|a+b|}=x,\sqrt[4]{|c+d|}=y\left(x,y\ge0\right)\)

Cần Cm \(x^4+y^4\ge2\left(x^3y-xy^3\right)\left(1\right)\)

<=> \(x^3\left(x-2y\right)+y^4+2xy^3\ge0\left(2\right)\)

+ Nếu \(x\ge2y\)=> BĐT được CM

+ Nếu \(x\le2y\)

(1) <=> \(x^4+y^4+2xy^3\ge2x^3y\)

Mà \(x^4+x^2y^2\ge2x^3y\)

=> Cần CM \(y^4+2xy^3-x^2y^2\ge0\)

<=> \(y^4+xy^2\left(2y-x\right)\ge0\)luôn đúng do \(x\le2y\)

=> BĐT được CM

Dấu bằng xảy ra khi a=b=c=d=0